Beweisen Sie Ein Viereck Ist Ein Parallelogramm Mit Slope In Forex

Bewertung von Mastery Guide Holt California Geometry Bewertung für Mastery Workbook Copyright by Holt, Rinehart und Winston. Alle Rechte vorbehalten. Kein Teil dieser Publikation darf ohne schriftliche Erlaubnis des Herausgebers in irgendeiner Form oder mit irgendwelchen Mitteln elektronisch oder mechanisch vervielfältigt oder übertragen werden, einschließlich Fotokopie, Aufzeichnung oder Informationsspeicherung und - wiederherstellung. Anfragen für die Genehmigung, Kopien von irgendeinem Teil der Arbeit zu machen, sollten an folgende Adresse geschickt werden: Permissions Department, Holt, Rinehart und Winston, 10801 N. MoPac Expressway, Gebäude 3, Austin, Texas 78759. HOLT und das Eulen-Design sind Marken Lizenziert an Holt, Rinehart und Winston, in den Vereinigten Staaten von Amerika und / oder anderen Ländern registriert. Gedruckt in den Vereinigten Staaten von Amerika Wenn Sie diese Materialien kostenlos als Prüfungskopien erhalten haben, behalten Holt, Rinehart und Winston das Eigentum an den Materialien und dürfen nicht weiterverkauft werden. Der Weiterverkauf von Kopien ist streng verboten. Der Besitz dieser Publikation im Druckformat berechtigt nicht, diese Publikation oder einen Teil davon in ein elektronisches Format zu konvertieren. ISBN 13: 978-0-03-099025-0 ISBN 10: 0-03-099025-4 1 2 3 4 5 6 7 8 9 862 10 09 08 07 Copyright von Holt, Rinehart und Winston. I i i Holt Geometry Alle Rechte vorbehalten. Inhaltsverzeichnis Kapitel 1. 1 Kapitel 2. 15 Kapitel 3. 29 Kapitel 4. Kapitel 5. Kapitel 6. 109 Kapitel 9. 109 Kapitel 10. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 13 Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 13 Kapitel 12. Kapitel 13 Kapitel 12. Kapitel 12. Kapitel 12. Kapitel 12. Von Holt, Rinehart und Winston. 1 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse 1-1 LESSON Review für Mastery Punkte, Linien und Ebenen verstehen Ein Punkt hat keine Größe. Es wird mit einem Großbuchstaben benannt. Alle nachfolgenden Figuren enthalten Punkte. Abbildung Charakteristik Diagramm Wörter und Symbole Zeile 0 Endpunkte erstreckt sich für immer in zwei Richtungen. Linie AB oder AB Linie Segment oder Segment 2 Endpunkte hat eine endliche Länge 8 9 Segment XY oder XY Strahl 1 Endpunkt erstreckt sich für immer in einer Richtung 1 2 Ray RQ oder RQ Ein Strahl wird mit seinem Endpunkt benannt. (Ebene FGH oder Ebene V Zeichnen und beschriften Sie für jede Figur ein Diagramm 1. Punkt W 2. Zeile MN 3. JK 4. EF Benennen Sie jede Zahl mit Hilfe von Wörtern und Symbolen 5. 6 4 3 7 Nennen Sie das Flugzeug auf zwei verschiedene Arten: 8. 7 8. - Punkt P Copyright durch Holt, Rinehart und Winston 2 Holt Geometrie Alle Rechte vorbehalten Name Datum Klasse 1-1 LESSON Begriff Bedeutung Kollineare Punkte, (F und G sind kollinear, F, G und H sind nichtkollineare nichtkollineare Punkte, die nicht auf der gleichen Linie liegen, koplanare Punkte oder Linien, die in derselben Ebene liegen 7 8 9 W, X und Y sind koplanar W, X, Y und Z sind nicht koplanare, nicht koplanare Punkte oder Linien, die nicht in der gleichen Ebene liegen Zahlen, die sich schneiden, teilen sich einen gemeinsamen Satz von Punkten. Im ersten Modell oben schneidet FH FG an Punkt F. In der zweiten Modell, XZ schneidet Flugzeug WXY am Punkt X. Verwenden Sie die Abbildung für Übungen 914. Nennen Sie jede der folgenden 0. 9. drei kollineare Punkte 10. drei nichtkollineare Punkte 11. vier koplanare Punkte 12. vier noncoplanar Punkte 13. zwei Linien, die Schneiden CD 14. die Kreuzung von JK und Ebene R Review für Mastery Verständnis Punkte, Linien und Ebenen weiter Copyright by Holt, Rinehart und Winston. 3 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON LESSON 1-2 Bewertung für Mastery Measuring and Constructing Segments Der Abstand zwischen zwei Punkten ist die Länge des Segments, das sie verbindet. Zentimeter (cm) 0 1 2 3 4 5 6 7 (Der Abstand zwischen E und J ist EJ, die Länge von EJ. Um die Entfernung zu finden, subtrahieren Sie die Zahlen, die den Punkten entsprechen, und nehmen Sie dann den Absolutwert EJ 7 1 6 6 cm Benutzen Sie die obige Abbildung, um die jeweilige Länge zu finden 1. EG 2. EF 3. FH 0 01 12 02 2 1 X Auf PR steht Q zwischen P und R. Bei PR 16 finden wir QR PQ QR PR 9 X 16 x 7 QR 7 4. Y 5. Z Finden Sie JK Finden Sie BC 6. 3 4 6 NN 7. 7 8 9 AA Finde X. 8. X 9. 3 4 5 JJ Finde DF ST. Copyright by Holt, Rinehart und Winston 4 Holt Geometry Alle Rechte vorbehalten Name Datum Klasse LESSON 1-2 Bewertung für Mastery Measuring and Constructing Segments continue Segmente sind kongruent, wenn ihre Längen gleich sind AB BC Die Länge von AB entspricht der Länge Von BC BC BC AB ist kongruent zu BC Kopieren einer Segmentmethode Schritte Skizze mit Schätzung Schätzung der Länge des Segments Skizzieren eines Segments, das etwa die gleiche Länge hat Ziehen mit einem Lineal Verwenden Sie ein Lineal, um die Länge des Segments zu messen . Verwenden Sie das Lineal, um ein Segment mit der gleichen Länge zu zeichnen. Konstrukt mit Kompass und Lineal Zeichnen Sie eine Linie und markieren Sie einen Punkt darauf. Öffnen Sie den Kompass auf die Länge des ursprünglichen Segments. Markieren Sie ein Segment auf Ihrer Linie auf der gleichen Länge. Siehe Dreieck ABC oben für Übungen 10 und 11. 10. Skizze LM, die kongruent zu AC ist. 11. Verwenden Sie ein Lineal, um XY zu ziehen, das kongruent ist. 12. Verwenden Sie einen Kompass, um ST zu konstruieren, der kongruent zu JK ist. Der Mittelpunkt eines Segments trennt das Segment in zwei kongruente Segmente. In der Figur ist P der Mittelpunkt von NQ. X X 0 1 13. PQ ist kongruent zu. 14. Was ist der Wert von x 15. Finden Sie NP, PQ und NQ. Copyright by Holt, Rinehart und Winston. 5 Holt Geometrie Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Messen und Konstruieren von Winkeln 1-3 Ein Winkel ist eine Figur, die aus zwei Strahlen oder Seiten besteht, die einen gemeinsamen Endpunkt haben, den Eckpunkt des Winkels. 8 9. Es gibt vier Möglichkeiten, diesen Winkel zu benennen. Y Verwenden Sie den Scheitelpunkt. XYZ oder ZYX Verwenden Sie den Scheitel und einen Punkt auf jeder Seite. 2 Verwenden Sie die Nummer. Benennen Sie jeden Winkel auf drei Arten. 1. 0 2 1 2. (3. Name drei verschiedene Winkel in der Abbildung Winkel akut rechts stumpf gerade Modell AAAA Mögliche Maßnahmen 0 a 90 a 90 90 a 180 a 180 Klassifizieren Sie jeden Winkel als akut, rechts, stumpf oder gerade 4. QMN 6. PMQ Der Scheitel ist Y. Die Seiten sind YX und YZ Copyright von Holt, Rinehart und Winston 6 Holt Geometry Alle Rechte vorbehalten Name Datum Klasse LESSON 1-3 Review Für Mastery Measuring and Constructing Angles Fortsetzung Sie können einen Winkelmesser verwenden, um das Maß eines Winkels zu finden 1OO 8O 11O 7O 1 2 OOO 1 8 O 5 O 1 4 O 4 O 1 5 O 8 O 1 OO 2 O 1 7 O 1 O 8 O 1 O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O O GEW Das Maß von XVU kann durch Addition gefunden werden 6 8 7 5 mXVU mXVW mWVU 48 48 96 Winkel sind kongruent, wenn ihre Maße gleich sind In der Figur XVW WVU, weil die Winkel gleiche Maße haben VW ist eine Winkelhalbierende von XVU, weil es XVU in zwei kongruente Winkel teilt. Finden Sie jedes Winkelmaß. . 9. mCFB, wenn AFC ein gerader Winkel ist. 10. mEFA, wenn der Winkel kongruent zu DFE ist. 11. mEFC wenn DFC AFB. 12. mCFG, wenn FG eine Winkelhalbierende von CFB ist. DEG ist akut. GEF ist stumpf. Copyright by Holt, Rinehart und Winston. 7 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Paare von Winkeln 1-4 Winkelpaare benachbarte Winkel Lineare Paare Vertikale Winkel haben denselben Scheitelpunkt und teilen sich eine gemeinsame Seite benachbarten Winkeln, deren nicht gebräuchliche Seiten gegenüberliegende, nicht benachbarte Winkel sind, die durch zwei sich schneidende Linien 1 und 2 gebildet werden . 3 und 4 sind benachbart und bilden ein lineares Paar. 5 und 6 sind vertikale Winkel. Sagen Sie, ob 7 und 8 in jeder Figur nur benachbart sind, benachbart sind und ein lineares Paar bilden oder nicht benachbart sind. 1. 2. 3. Sagen Sie, ob die angegebenen Winkel nur benachbart sind, benachbart sind und ein lineares Paar bilden oder nicht benachbart sind. 4. 5 und 4 5. 1 und 4 6. 2 und 3 Nennen Sie jede der folgenden Angaben. 7. ein Paar vertikaler Winkel 8. ein lineares Paar 9. ein Winkel neben 4 Copyright durch Holt, Rinehart und Winston. 8 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON 1-4 Bewertung für Mastery Paare von Winkeln Fortsetzung Winkelpaare Ergänzende Winkel Zusatzwinkel Die Summe der Winkelmaße beträgt 90 Summe der Winkelmaße beträgt 180 m1 m2 90 In jedem Paar sind 1 und 2 komplementär. M3 m4 180 In jedem Paar sind 3 und 4 ergänzend. Sagen Sie, ob jedes Paar von markierten Winkeln komplementär, ergänzend oder keines ist. 10. 11. Finden Sie das Maß für jeden der folgenden Winkel. 12. Ergänzung von S 3 13. Ergänzung von S 14. Komplement von R 2 15. Ergänzung von R 16. LMN und UVW sind komplementär. Finden Sie das Maß für jeden Winkel, wenn mLMN (3x 5) und mUVW 2x. Copyright by Holt, Rinehart und Winston. 9 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Verwenden von Formeln in Geometrie 1-5 Der Umfang einer Figur ist die Summe der Längen der Seiten. Der Bereich ist die Anzahl der quadratischen Einheiten, die in der Figur eingeschlossen sind. Abbildung Rechteck Quadratisches Modell W W S S S S Perimeter P 2 2w oder 2 (w) P 4s Fläche A w A s 2 Finden Sie den Umfang und die Fläche jeder Figur. 1. Rechteck mit 4 ft, w 1 ft 2. Quadrat mit s 8 mm 3. CM 4. IN IN X X Der Umfang eines Dreiecks ist die Summe seiner Seitenlängen. Die Basis und die Höhe werden verwendet, um das Gebiet zu finden. H A C B B C H A Perimeterbereich P a b c A 1 2 bh oder bh 2 Finden Sie den Umfang und die Fläche jedes Dreiecks. 5. FT FT YFT 6. 9 cm 6,7 cm 6 cm 8,5 cm Copyright by Holt, Rinehart und Winston. 10 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON 1-5 Bewertung für Mastery Verwendung von Formeln in der Geometrie Fortsetzung Circles Circumference Area Models DR Wörter pi mal den Durchmesser oder 2 mal pi mal den Radius pi mal dem Quadrat des Radius Formeln C d oder C 2r A r 2 MC 2r A r 2 C 2 (4) A (4) 2 C 8 A 16 C 25.1 m A 50.3 m 2 Finden Sie den Umfang und die Fläche jedes Kreises. Verwenden Sie die Taste auf Ihrem Rechner. Runde bis zum nächsten Zehntel. 7. Kreis mit einem Radius von 11 Zoll 8. Kreis mit einem Durchmesser von 15 Millimeter 9. IN 10. CM 11. M 12. MM Abstand um den Kreisraum innerhalb des Kreises Copyright by Holt, Rinehart und Winston. 11 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON 1-6 Bewertung für Mastery-Mittelpunkt und Distanz in der Koordinatenebene Der Mittelpunkt eines Liniensegments trennt das Segment in zwei Hälften. Sie können die Mittelpunktformel verwenden, um den Mittelpunkt des Segments mit den Endpunkten G (1, 2) und H (7, 6) zu finden. 7 XY 7 0 - (4, 4) (1, 2) ((7, 6) M x 1 x 2 2. y 1 y 2 2 M 1 7 2. 2 6 2 M 8 2. 8 2 M (4 , 4) Finden Sie die Koordinaten des Mittelpunkts jedes Segments 1. 6 XY 0 6 3 (4, 5) (2, 5) 2. 3 XY 0 3 3 3 4 (1, 4) 3 (3, 2) 3. QR mit den Endpunkten Q (0, 5) und R (6, 7) 4. JK mit den Endpunkten J (1, 4) und K (9, 3) Angenommen, M (3, 1) ist der Mittelpunkt von CD und C (3, 1) M x 1 x 2 2. y 1 y 2 2 x-Koordinate der D y - Koordinate von D 3 ist, können Sie die Mittelpunktformel verwenden X 1 x 2 2 Stellen Sie die Koordinaten gleich ein 1 y 1 y 2 2 3 1 x 2 2 Ersetzen Sie (x 1. y 1) mit (1, 4) 1 4 y 2 2 6 1 x 2 Multiplizieren Sie beide Seiten mit 2 2 4 y 2 5 x 2 Subtrahieren, um für x 2 und y 2 zu lösen. 6 y 2 Die Koordinaten von D sind (5, 6) (6, 0) Was sind die Koordinaten von S 6. M (7, 1) ist der Mittelpunkt von WX und X hat Koordinaten (1, 5) Was sind die Koordinaten von WM ist der Mittelpunkt von HG Holt, Rinehart und Winston. 12 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON 1-6 Bewertung für Mastery-Mittelpunkt und Entfernung in der Koordinatenebene Fortsetzung Die Distance Formula kann verwendet werden, um den Abstand d 7 XY 7 0 (1, 2) (7, 6) D zwischen den Punkten A und B in zu finden Die Koordinatenebene. D (x 2 x 1) 2 (y 2 y 1) 2 (7 1) 2 (6 2) 2 (x 1 y 1) (1, 2) (x 2 y 2) (7,6) 2 4 2 Subtrahieren. 36 16 Platz 6 und 4. 52 Hinzufügen. 7.2 Verwenden Sie einen Taschenrechner. Verwenden Sie die Distance Formula, um die Länge jedes Segments oder die Distanz zwischen den einzelnen Punkten zu finden. Runde bis zum nächsten Zehntel. 7. QF mit den Endpunkten Q (2, 4) und R (3, 9) 8. EF mit den Endpunkten E (8, 1) und F (1, 1) 9. T (8, 3) und U (5, 5 ) 10. N (4, 2) und P (7, 1) Sie können auch das Pythagoreische Theorem verwenden, um Abstände in der Koordinatenebene zu finden. Finden Sie den Abstand zwischen J und K. c 2 a 2 b 2 Pythagoreisches Theorem X Y C B A 5 2 6 2 a 5 Einheiten und b 6 Einheiten 25 36 Quadrat 5 und 6 61. C 61 oder etwa 7.8 Nehmen Sie die Quadratwurzel. Verwenden Sie das pythagoreische Theorem, um die Entfernung, auf die nächste Zehnte, zwischen jedem Paar von Punkten zu finden. 11. 6 X Y 0 6 3: (4, 5) 9 (0, 1) 12. X Y -. Der Abstand d zwischen den Punkten A und B ist die Länge von AB. Seite b ist 6 Einheiten. Seite a ist 5 Einheiten. Copyright by Holt, Rinehart und Winston. 13 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery-Transformationen in der Koordinatenebene 1-7 In einer Transformation wird jeder Punkt einer Figur an eine neue Position verschoben. Reflexionsdrehung Übersetzung. ABC ABC. . JKL JKL 2 3 4 2 3 4 RST RST Eine Figur wird über eine Linie gespiegelt. Eine Figur wird um einen festen Punkt gedreht. Eine Figur wird ohne Drehen auf eine neue Position geschoben. Identifizieren Sie jede Transformation. Verwenden Sie dann Pfeil-Notation, um die Transformation zu beschreiben. 1. Holt, Rinehart und Winston 4. Holt Geometry Alle Rechte vorbehalten Name Datum Klasse LESSON 1-7 Bewertung für Mastery Transformations in der Koordinatenebene fortgesetzt Das Dreieck QRS hat bei Q (4, 1), R (3, 4), XY 2 1 2 3 3 1 und S (0, 0) Scheitelpunkte (1, 4), R (4, 3) und S (0, 0) dar. Die Transformation ist eine Drehung, die durch eine Regel wie (x, y) (x 4, y 1) beschrieben werden kann. (0, 1) S (0 4, 1 1) S (4, 0) T (2, 1) R (3, 5) Das Dreieck HJK hat Ecken zu H (3, 1), XYJ (3, 4) und K (2, Nach einer Transformation hat das Bild der Abbildung Eckpunkte bei H (1, 3), J (1, 2) und K (4, 2). 6. Dreieck CDE hat Ecken bei C (4, Nach einer Transformation hat das Bild der Figur Scheitelpunkte bei C (4, 6), D (1, 6) und E (2, 1) ). Finden Sie die Koordinaten für jedes Bild nach der angegebenen Übersetzung. 7. Präimage: XYZ bei X (6, 1), Y (4, 0), Z (1, 3) Vorschrift: (FGH bei F (9, (X, y) (x 3, y 1) 9. Vorbild: BCD bei B (0, 2), C (7, 1), D (1, 5) Regel: (x, y) (x 7, y 1) Copyright von Holt, Rinehart und Winston. 15 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Beherrschung mit Induktivität, um Vermutungen zu treffen 2-1 Wenn Sie eine allgemeine Regel oder Schlussfolgerung basierend auf einem Muster machen, verwenden Sie induktives Denken. Eine auf einem Muster basierende Schlussfolgerung wird als Vermutung bezeichnet. Pattern Conjecture Next Zwei Items 8, 3, 2, 7. Jeder Begriff ist 5 mehr als der vorherige Begriff. 7 5 12 12 5 17 45 Das Maß jedes Winkels ist das halbe Maß des vorherigen Winkels. 22.5 11.25 Suchen Sie das nächste Element in jedem Muster. 1. 1 4. 1 2. 3 4. 1. 2. 100, 81, 64, 49. 3. 3 6 10 4. Jede Vermutung beenden. 5. Wenn die Seitenlänge eines Quadrats verdoppelt wird, ist der Umfang des Quadrats. Die Anzahl der nicht überlappenden Winkel, die durch n Linien gebildet werden, die sich in einem Punkt schneiden, ist. Verwenden Sie die Abbildung, um die Vermutung in Übung 7 zu vervollständigen. 7. Der Umfang einer Zahl, die n dieser Dreiecke hat 1 1 1 1 0 3 1 1 1 1 1 1 0 4 1 1 1 0 5 1 1 1 1 0 6 Ist. Copyright by Holt, Rinehart und Winston. 16 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery mit Induktivität, um Vermutungen zu machen 2-1 Da eine Vermutung eine gebildete Vermutung ist, kann sie wahr oder falsch sein. Es braucht nur ein Beispiel oder Gegenbeispiel, um zu beweisen, daß eine Vermutung falsch ist. Vermutung: Für jede ganze Zahl n, n 4n. N n 4n Wahr oder Falsch 3 3 4 (3) 3 12 true 0 0 4 (0) 0 0 true 2 2 4 (2) 2 8 false n 2 ist ein Gegenbeispiel, daher ist die Vermutung falsch. Zeigen Sie, dass jede Vermutung falsch ist, indem Sie ein Gegenbeispiel finden. 8. Wenn drei Linien in der gleichen Ebene liegen, schneiden sie sich in mindestens einem Punkt. Die Punkte A, G und N sind kollinear. Wenn AG 7 Zoll und GN 5 Zoll, dann AN 12 Zoll. 10. Für alle reellen Zahlen x und y, wenn xy, dann x 2 y 2. 11. Die Gesamtzahl der Winkel in der Figur ist 3. 12. Sind zwei Winkel scharf, so ist die Summe ihrer Maße gleich dem Maß von Einen stumpfen Winkel. Bestimmen Sie, ob jede Vermutung wahr ist. Wenn nicht, schreiben oder zeichnen Sie ein Gegenbeispiel. 13. Punkte Q und R sind kollinear. 14. Wenn J zwischen H und K liegt, dann HJ JK. Copyright by Holt, Rinehart und Winston. 17 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Eine Bedingungsanweisung ist eine Anweisung, die als if-then-Anweisung geschrieben werden kann, wenn p, dann q. Wenn Sie dieses Handy kaufen, dann erhalten Sie 10 kostenlose Klingelton Downloads. Manchmal ist es notwendig, eine bedingte Anweisung neu zu schreiben, so dass sie sich in if-then-Form befindet. Voraussetzung: Eine Person, die Putting praktiziert wird ihr Golfspiel verbessern. If-Then Form: Wenn eine Person übt Putting, dann wird sie verbessern ihre Golf-Spiel. Eine bedingte Anweisung hat einen falschen Wahrheitswert nur, wenn die Hypothese (H) wahr ist und die Schlussfolgerung (C) falsch ist. Für jede Bedingung, unterstreichen die Hypothese und doppelt unterstreichen die Schlussfolgerung. 1. Wenn x eine gerade Zahl ist, dann ist x durch 2 teilbar. 2. Der Umfang eines Kreises ist 5 Zoll, wenn der Durchmesser des Kreises 5 Zoll ist. 3. Wenn eine Gerade mit den Punkten J, K und L in der Ebene P liegt, so sind J, K und L koplanar. Für Übungen 46 schreiben Sie eine bedingte Anweisung aus jeder gegebenen Anweisung. 4. Die kongruenten Segmente haben gleiche Maßnahmen. 5. Am Dienstag ist Spielpraxis um 6:00 Uhr. 6. Benachbarte Winkel Lineare Paare Bestimmen Sie, ob die folgende Bedingung wahr ist. Wenn false, geben Sie ein Gegenbeispiel. 7. Sind zwei Winkel ergänzend, so bilden sie ein lineares Paar. Die Hypothese kommt nach dem Wort if. Die Schlussfolgerung kommt nach dem Wort dann. Bewertung für Mastery Conditional Statements 2-2 Copyright by Holt, Rinehart und Winston. 18 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON C H Review für Mastery Bedingte Anweisungen Fortsetzung 2-2 Die Negation einer Aussage, nicht p, hat den entgegengesetzten Wahrheitswert der ursprünglichen Aussage. Ist p wahr, so ist p nicht falsch. Ist p falsch, so ist p nicht wahr. Statement Beispiel Wahrheitswert Bedingung Wenn eine Figur ein Quadrat ist, hat sie vier rechte Winkel. True Converse: Schalter H und C. Wenn eine Figur vier rechte Winkel hat, dann ist sie ein Quadrat. Falsche Inverse: Negieren Sie H und C. Wenn eine Figur kein Quadrat ist, dann hat sie nicht vier rechte Winkel. Falsch Contrapositive: Schalten und negieren H und C. Wenn eine Figur nicht vier rechte Winkel hat, dann ist sie kein Quadrat. True Schreiben Sie die Umkehrung, umgekehrt und kontrapositive jeder bedingten Anweisung. Finden Sie den Wahrheitswert von jedem. 8. Wenn ein Tier ein Gürteltier ist, dann ist es nachtaktiv. 9. Wenn y 1, dann y 2 1. 10. Wenn ein Winkel ein Maß kleiner als 90 hat, dann ist es akut. Copyright by Holt, Rinehart und Winston. 19 Holt Geometry Alle Rechte vorbehalten. Name Datum Unterricht. Review für Mastery mit deduktiven Reasoning zu prüfen, Conjectures Mit induktiven Argumentation, verwenden Sie Beispiele, um eine Vermutung zu machen. Mit deduktivem Denken verwenden Sie Tatsachen, Definitionen und Eigenschaften, um Schlussfolgerungen zu ziehen und zu beweisen, dass Vermutungen wahr sind. Gegeben: Wenn zwei Punkte in einer Ebene liegen, dann liegt die Linie, die diese Punkte enthält, auch in der Ebene. A und B liegen in der Ebene N. Vermutung: AB liegt in Ebene N. Eine gültige Form der deduktiven Argumentation, die Sie Schlussfolgerungen aus wahren Tatsachen ziehen lässt, nennt man das Gesetz der Detachment. Wenn Sie 2 haben, dann können Sie einen Snack kaufen. Sie haben 2. Wenn Sie 2 haben, dann können Sie einen Snack kaufen. Sie können einen Snack kaufen. Vermutung Sie können einen Imbiss kaufen. Sie haben 2. Valid Conjecture Ja die Bedingung ist wahr und die Hypothese ist wahr. Nein, die Hypothese kann oder kann nicht wahr sein. Zum Beispiel, wenn Sie Geld geliehen, könnten Sie auch einen Snack kaufen. Sagen Sie, ob jede Schlussfolgerung induktiv oder deduktiv betrachtet. 1. Ein Schild in der Cafeteria sagt, dass eine Autowaschanlage am letzten Samstag im Mai stattfindet. Morgen ist der letzte Samstag im Mai, deshalb kommt Justin zu dem Schluss, dass die Autowäsche morgen ist. 2. Bis zu Beginn jeder lateinischen Klasse hat der Lehrer die Schüler das Vokabular überprüft. Die lateinische Klasse beginnt, und Jamilla geht davon aus, dass sie zuerst das Vokabular überprüfen. 3. Gegenüberliegende Strahlen sind zwei Strahlen, die einen gemeinsamen Endpunkt haben und eine Linie bilden. YX und YZ sind entgegengesetzte Strahlen. 8. Bestimmen Sie, ob jede Vermutung gültig ist durch das Gesetz der Detachment. 4. Gegeben: Wenn Sie die Titan-Achterbahn in Arlington, Texas fahren, dann werden Sie 255 Fuß fallen. Michael fuhr die Achterbahn Titan. Vermutung: Michael fiel 255 Fuß. 5. Gegeben: Ein Segment, das einen Kreisdurchmesser hat, hat Endpunkte auf dem Kreis. GH hat Endpunkte auf einem Kreis. Vermutung: GH ist ein Durchmesser. 2-3 Copyright von Holt, Rinehart und Winston. 20 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Beherrschung mit Deduktiven Reasoning zu bestätigen Conjectures Fortsetzung Eine weitere gültige Form der deduktiven Argumentation ist das Gesetz des Syllogismus. Sie ist der transitiven Eigenschaft der Gleichheit ähnlich. Transitive Eigenschaft der Gleichheit Gesetz der Syllogismus Wenn y 10x und 10x 20, dann y 20. Gegeben: Wenn Sie ein Pferd haben, dann müssen Sie es zu füttern. Wenn Sie ein Pferd füttern müssen, müssen Sie jeden Morgen früh aufstehen. Vermutung: Wenn du ein Pferd hast, musst du jeden Morgen früh aufstehen. Bestimmen Sie, ob jede Vermutung durch das Gesetz des Syllogismus gültig ist. 6. Gegeben: Wenn Sie ein Auto kaufen, dann können Sie zur Schule fahren. Wenn Sie zur Schule fahren können, dann fahren Sie nicht mit dem Bus. Vermutung: Wenn Sie ein Auto kaufen, dann werden Sie nicht mit dem Bus fahren. 7. Gegeben: Wenn K stumpf ist, dann hat es kein Maß von 90. Wenn ein Winkel nicht ein Maß von 90 hat, dann ist es nicht ein rechter Winkel. Vermutung: Ist K stumpf, so ist es kein rechter Winkel. 8. Gegeben: Sind zwei Segmente kongruent, so haben sie das gleiche Maß. Wenn zwei Segmente jeweils ein Maß von 6,5 Zentimetern haben, dann sind sie kongruent. Vermutung: Sind zwei Segmente kongruent, so haben sie jeweils ein Maß von 6,5 Zentimetern. Aus den gegebenen Informationen eine Schlussfolgerung ziehen. 9. Wenn LMN in die Koordinatenebene übersetzt wird, hat es die gleiche Größe und Form wie sein Vorbild. Wenn ein Bild und ein Vorbild die gleiche Größe und Form haben, haben die Figuren gleiche Umgrenzungen. LMN wird in die Koordinatenebene übersetzt. 10. Sind R und S komplementär zu demselben Winkel, so sind die beiden Winkel kongruent. Sind zwei Winkel kongruent, so sind sie ergänzend zu demselben Winkel. R und S komplementär zu demselben Winkel sind. 2-3 Copyright von Holt, Rinehart und Winston. 21 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON p q q p p q Review für Mastery Biconditionale Anweisungen und Definitionen 2-4 Eine biconditionale Anweisung kombiniert eine bedingte Anweisung, wenn p, dann q mit ihrer Umkehrung, wenn q, dann p. Bedingung: Sind die Seiten eines Dreiecks kongruent, so sind die Winkel kongruent. Converse: Sind die Winkel eines Dreiecks kongruent, so sind die Seiten kongruent. Biconditional: Die Seiten eines Dreiecks sind nur dann kongruent, wenn die Winkel kongruent sind. Schreiben Sie die bedingte Anweisung und konversieren in jedem biconditional. 1. Lindsay nimmt Fotos für das Jahrbuch, wenn und nur wenn sie nicht Fußball spielt. 2. mABC mCBD genau dann, wenn BC eine Winkelhalbierende von ABD ist. Für jede Bedingung schreiben Sie die Umkehrung und eine biconditionale Anweisung. 3. Wenn Sie 6 Songs für 5,94 herunterladen können, kostet jeder Song 0,99. 4. Wenn eine Figur 10 Seiten hat, dann ist sie ein Dezagon. Copyright by Holt, Rinehart und Winston. 22 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Biconditionale Anweisungen und Definitionen Fortsetzung 2-4 Eine biconditionale Anweisung ist false, wenn entweder die bedingte Anweisung false ist oder ihre Umkehrung false ist. Der Mittelpunkt von QR ist genau dann M (3, 3), wenn die Endpunkte Q (6, 1) und R (0, 5) sind. Bedingung: Wenn der Mittelpunkt von QR M (3, 3) ist, dann sind die 0 XY 3 - (3, 3) 1 (6, 1) 2 (0, 5) 3 Endpunkte Q (6, 1) und R 0, 5). False Converse: Wenn die Endpunkte von QR Q (6, 1) und R (0, 5) sind, dann ist der Mittelpunkt von QR M (3, 3). True Die Bedingung ist falsch, da die Endpunkte von QR Q (3, 6) und R (3, 0) sein können. Also ist die biconditionale Aussage falsch. Definitionen können als biconditionals geschrieben werden. Definition: Umfang ist der Abstand um einen Kreis. Biconditional: Ein Maß ist der Umfang genau dann, wenn es der Abstand um einen Kreis ist. Bestimmen Sie, ob jedes biconditional wahr ist. Wenn false, geben Sie ein Gegenbeispiel. 5. Die Schüler spielen während der Halbzeit bei den Fußballspielen, wenn sie in der High School Band sind. 6. Ein Winkel in einem Dreieck misst 90 genau dann, wenn das Dreieck ein rechtes Dreieck ist. 7. a 4 und b 3 genau dann, wenn ab 12. Jede Definition als biconditional schreiben. Ein gleichschenkliges Dreieck weist mindestens zwei kongruente Seiten auf. 9. Deduktive Begründung erfordert die Verwendung von Tatsachen, Definitionen und Eigenschaften, um Schlussfolgerungen zu ziehen. Copyright by Holt, Rinehart und Winston. 23 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Algebraischer Beweis Ein Beweis ist ein logisches Argument, dass eine Schlussfolgerung wahr ist. Ein algebraischer Beweis verwendet algebraische Eigenschaften, einschließlich der verteilenden Eigenschaft und der Eigenschaften der Gleichheit. Eigenschaften von Gleichheit Symbole Beispiele Addition Wenn a b, dann a c b c. Wenn x 4, dann x 4 4 4. Subtraktion Wenn a b, dann a c b c. Wenn r 1 7, dann r 1 1 7 1. Multiplikation Wenn a b, dann ac bc. Wenn k 2 8, dann k 2 (2) 8 (2). Division Wenn a 2 und c 0, dann a c b c. Wenn 6 3t, dann 6 3 3t 3. Reflexive a a 15 15 Symmetrisch Wenn a b, dann b a. Wenn n 2, dann 2 n. Transitiv Wenn a b und b c, dann ein c. Wenn y 3 2 und 3 2 9, dann y 9. Substitution Wenn a b, so kann b in jedem beliebigen Ausdruck substituiert werden. Wenn x 7, dann 2x 2 (7). Wenn Sie eine algebraische Gleichung lösen, rechtfertigen Sie jeden Schritt, indem Sie eine Definition, eine Eigenschaft oder ein Stück der gegebenen Information verwenden. 2 (a 1) 6 Gegeben nach Gleichung 2a 2 6 Verteilende Eigenschaft 2 2 Subtraktion Gleichheitsgleichung 2a 8 Vereinfachen. 2a 2 8 2 Division Eigentum der Gleichheit a 4 Vereinfachen. Lösen Sie jede Gleichung. Schreibe eine Berechtigung für jeden Schritt. 1. n 6 3 10 2. 5 x 2x 3. y 4 7 3 4. 4 (t 3) 20 2-5 Copyright durch Holt, Rinehart und Winston. 24 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Algebraischer Beweis Fortsetzung 2-5 Wenn Sie algebraische Beweise in der Geometrie schreiben, können Sie auch Definitionen, Postulate, Eigenschaften und Stücke von gegebenen Informationen verwenden, um die Schritte zu rechtfertigen. MJKM mMKL Definition der kongruenten Winkel - X X. (5x 12) 4x Substitution Eigenschaft der Gleichheit x 12 0 Subtraktion Eigenschaft der Gleichheit x 12 Addition Eigenschaft der Gleichheit Eigenschaften der Kongruenz Symbole Beispiele Reflexive Abbildung A Abbildung A CDE CDE Symmetrisch Wenn Abbildung A Abbildung B , Dann Abbildung B Abbildung A. Wenn JK LM. Dann LM JK. Transitive Wenn Abbildung A Abbildung B und Abbildung B Abbildung C, dann Abbildung A Abbildung C. Wenn N P und P Q, dann N Q. Schreibe eine Berechtigung für jeden Schritt. 5. CE CD DE 3X 7 8 6X 6x 8 (3x 7) 6x 15 3x 3x 15 x 5 6. mPQR mPQS mSQR XX 2 3 0 1 90 2x (4x 12) 90 6x 12 102 6x 17 x Identifizieren Sie die Eigenschaft, die es rechtfertigt jede Aussage. 7. Wenn ABC DEF, dann DEF ABC. 8. 1 2 und 2 3, so 1 3. 9. Wenn FG HJ, dann HJ FG. 10. WX WX Copyright durch Holt, Rinehart und Winston. 25 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Hypothese Deduktive Begründung Definitionen Eigenschaften Postulates Theorems Fazit Review für Mastery Geometrischer Beweis 2-6 Um einen geometrischen Beweis zu schreiben, beginnen Sie mit der Hypothese einer bedingten. Wenden Sie deduktive Argumentation an. Beweisen Sie, dass der Abschluss der Bedingung wahr ist. Bedingung: Wenn BD die Winkelhalbierende von ABC und ABD 1 ist, dann DBC 1. Gegeben: BD ist die Winkelhalbierende von ABC und ABD 1. 1. Beweis: DBC 1 Beweis: 1. BD ist die Winkelhalbierende von ABC. 1. Gegeben 2. ABD DBC 2. Def. Der Winkelhalbierenden 3. ABD 1 3. Gegeben 4. DBC 1 4. Transitive Prop. Von 1. Gegeben: N ist der Mittelpunkt des MP. Q ist der Mittelpunkt von RP. Und PQ NM. 0 1 2 -. Beweisen: PN QR Schreibe eine Berechtigung für jeden Schritt. Beweis: 1. N ist der Mittelpunkt von MP. 1. 2. Q ist der Mittelpunkt von RP. 2. 3. PN NM 3. 4. PQ NM 4. 5. PN PQ 5. 6. PQ QR 6. 7. PN QR 7. Copyright durch Holt, Rinehart und Winston. 26 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Geometrischer Beweis Fortsetzung 2-6 Ein Theorem ist eine Aussage, die Sie nachweisen können. Sie können zweispaltige Beweise und deduktive Argumentation zu beweisen Theoreme. Congruent Supplements Theorem Sind zwei Winkel zu demselben Winkel (oder zu zwei kongruenten Winkeln) ergänzt, so sind die beiden Winkel kongruent. Rechtwinkliger Kongruenztheorem Alle rechten Winkel sind kongruent. Hier ist ein zweispaltiger Beweis eines Falles des Congruent Supplements Theorem. Gegeben sind: 4 und 5 sind ergänzend und 5 und 6 sind ergänzend. 4 6 5 7 Beweis: 4 6 Nachweis: Aussagen Gründe 1. 4 und 5 sind ergänzend. 1. Die 2. 5 und 6 sind ergänzend. 2. Gegeben 3. m4 m5 180 3. Definition zusätzlicher Winkel 4. m5 m6 180 4. Definition zusätzlicher Winkel 5. m4 m5 m5 m6 5. Substitution Eigenschaft der Gleichheit 6. m4 m6 6. Subtraktion Eigenschaft der Gleichheit 7. 4 6 7. Definition der kongruenten Winkel Füllen Sie die Lücken aus, um den zweispaltigen Beweis 1 2 des rechtwinkligen Kongruenztheorems abzuschließen. 2. Gegeben: 1 und 2 sind rechtwinklig. Beweis: 1 2 Beweis: Aussagen Gründe 1. a. 1. Gegeben 2. m1 90 2. b. 3. c. 3. Definition des rechten Winkels 4. m1 m2 4. d. 5. e. 5. Definition der kongruenten Winkel Copyright by Holt, Rinehart und Winston. 27 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery-Flussdiagramm und Absatz-Beweise 2-7 Zusätzlich zum zweispaltigen Beweis gibt es andere Arten von Beweisen, die Sie verwenden können, um Vermutungen wahr zu machen. Flowchart Proof Verwendet Felder und Pfeile. Schritte von links nach rechts oder von oben nach unten, wie durch Pfeile gezeigt. Die Rechtfertigung für jeden Schritt wird unter dem Feld geschrieben. Sie können einen Flussdiagrammbeweis des rechtwinkligen Kongruenztheorems schreiben. Gegeben: 1 und 2 sind rechtwinklig. 1 2 Beweisen: 1 2 1 und 2 sind rt. Gegeben m1 90, m2 90 Def. Von rt. M1 Transm. Prop. Von 1 2 Def. Von 1. Verwenden Sie den angegebenen zweispaltigen Beweis, um einen Flussdiagrammbeweis zu schreiben. Gegeben: V ist der Mittelpunkt von SW. Und W der Mittelpunkt von VT ist. 3 6 7 4 Beweis: SV WT Zwei-Säulen-Beweis: Aussagen Gründe 1. V ist der Mittelpunkt der SW. 1. Gegeben 2. W ist der Mittelpunkt von VT. 2. Angesichts 3. SV VW. VW WT 3. Definition des Mittelpunkts 4. SV WT 4. Transitives Eigentum der Gleichheit Copyright by Holt, Rinehart und Winston. 28 Holt Geometry Alle Rechte vorbehalten. Name Datum Klasse LESSON Review für Mastery Flussdiagramm und Absatz Proofs Fortsetzung 2-7 Um einen Absatznachweis zu schreiben, verwenden Sie Sätze, um einen Absatz zu schreiben, der die Aussagen und Gründe enthält. Sie können den angegebenen zweispaltigen Beweis verwenden, um einen Absatzbeweis zu schreiben. Gegeben: AB BC und BC DE. Beweis: AB DE Zwei-Säulen-Beweis: Aussagen Gründe 1. AB BC. BC DE 1. Gegeben 2. AB BC, BC DE 2. Definition der kongruenten Segmente 3. AB DE 3. Transitives Gleichheitsgleichgewicht 4. AB DE 4. Definition der kongruenten Segmente Paragraph Beweis: Es ist gegeben, dass AB BC und BC DE . So AB BC und BC DE durch die Definition der kongruenten Segmente. By the Transitive Property of Equality, AB DE. Thus, by the definition of congruent segments, AB DE. 2. Use the given two-column proof to write a paragraph proof. Given: JKL is a right angle. 1 2 . Prove: 1 and 2 are complementary angles. Two-Column Proof: Statements Reasons 1. JKL is a right angle. 1. Given 2. mJKL 90 2. Definition of right angle 3. mJKL m1 m2 3. Angle Addition Postulate 4. 90 m1 m2 4. Substitution 5. 1 and 2 are complementary angles. 5. Definition of complementary angles Paragraph Proof: Copyright by Holt, Rinehart and Winston. 29 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Lines and Angles 3-1 Lines Description Examples parallel lines that lie in the same plane and do not intersect symbol: M K perpendicular lines that form 90 angles symbol: skew lines that do not lie in the same plane and do not intersect Parallel planes are planes that do not intersect. For example, the top and bottom of a cube represent parallel planes. Use the figure for Exercises 13. Identify each of the following. 1. a pair of parallel lines J G H 2. a pair of skew lines 3. a pair of perpendicular lines Use the figure f or Exercises 4 9. Identify each of the following. ( 4. a segment that is parallel to DG 5. a segment that is perpendicular to GH 6. a segment that is skew to JF 7. one pair of parallel planes 8. one pair of perpendicular segments, 9. one pair of skew segments, not including GH not including JF m k k and m are skew. Copyright by Holt, Rinehart and Winston. 30 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Lines and Angles continued 3-1 A transversal is a line that intersects two lines in a plane at different points. Eight angles are formed. Line t is a transversal of lines a and b. T A 1 2 3 4 5 6 7 8 B Angle Pairs Formed by a Transversal Angles Description Examples corresponding angles that lie on the same side of the transversal and on the same sides of the other two lines T A 4 8 B alternate interior angles that lie on opposite sides of the transversal, between the other two lines T A 4 5 B alternate exterior angles that lie on opposite sides of the transversal, outside the other two lines T A 2 7 B same-side interior angles that lie on the same side of the transversal, between the other two lines also called consecutive interior angles T A 4 6 B Use the figure for Exercises 1013. Give an example of each type of angle pair. 1 2 3 4 5 6 7 8 10. corresponding angles 11. alternate exterior angles 12. same-side interior angles 13. alternate interior angles Use the figure for Exercises 1416. Identify the transversal and classify each angle pair. 1 2 3 4 M N P 14. 1 and 2 15. 2 and 4 16. 3 and 4 Copyright by Holt, Rinehart and Winston. 31 Holt Geometry All rights reserved. Name Date Class LESSON 3-2 Review for Mastery Angles Formed by Parallel Lines and Transversals According to the Corresponding Angles Postulate, if two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent. R S T Determine whether each pair of angles is congruent according to the Corresponding Angles Postulate. 1 2 3 4 1. 1 and 2 2. 3 and 4 Find each angle measure. 1 67 142 ( X 3. m1 4. mHJK X X. X X - 0 1. 5. mABC 6. mMPQ 1 3 2 4 Copyright by Holt, Rinehart and Winston. 32 Holt Geometry All rights reserved. Name Date Class LESSON 3-2 Review for Mastery Angles Formed by Parallel Lines and Transversals continued If two parallel lines are cut by a transversal, then the following pairs of angles are also congruent. Angle Pairs Hypothesis Conclusion alternate interior angles 2 6 3 7 C T D 2 3 6 7 alternate exterior angles 1 4 8 5 Q T R 1 4 5 8 If two parallel lines are cut by a transversal, then the pairs of same-side interior angles are supplementary. Find each angle measure. 3 111 4 7. m3 8. m4 138 X 2 3 4 A A - 0. 9. mRST 10. mMNP Y Y 7. 8 N N. 11. mWXZ 12. mABC m5 m6 180 m1 m2 180 Copyright by Holt, Rinehart and Winston. 33 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Proving Lines Parallel 3-3 Converse of the Corresponding Angles Postulate If two coplanar lines are cut by a transversal so that a pair of corresponding angles are congruent, then the two lines are parallel. You can use the Converse of the Corresponding Angles Postulate to show that two lines are parallel. 1 2 Q R 3 4 Given: 1 3 1 3 1 3 are corresponding angles. q r Converse of the Corresponding Angles Postulate Given: m2 3x, m4 (x 50), x 25 m2 3(25) 75 Substitute 25 for x. m4 (25 50) 75 Substitute 25 for x. m2 m4 Transitive Property of Equality 2 4 Definition of congruent angles q r Converse of the Corresponding Angles Postulate For Exercises 1 and 2, use the Converse of the Corresponding Angles Postulate and the given information to show that c d. 1. Given: 2 4 2. Given: m1 2x, m3 (3x 31), x 31 1 2 D C 3 4 Copyright by Holt, Rinehart and Winston. 34 Holt Geometry All rights reserved. Name Date Class LESSON 3-3 Review for Mastery Proving Lines Parallel continued You can also prove that two lines are parallel by using the converse of any of the other theorems that you learned in Lesson 3-2. Theorem Hypothesis Conclusion Converse of the Alternate Interior Angles Theorem 2 A T B 3 2 3 a b Converse of the Alternate Exterior Angles Theorem 4 F T G 1 1 4 f g Converse of the Same-Side Interior Angles Theorem 1 S T 2 m1 m2 180 s t For Exercises 35, use the theorems and the given information to show that j k. 3. Given: 4 5 4. Given: m3 12x, m5 18x, x 6 5. Given: m2 8x, m7 (7x 9), x 9 J K 1 2 3 4 5 6 7 8 Copyright by Holt, Rinehart and Winston. 35 Holt Geometry All rights reserved. Name Date Class LESSON 3-4 Review for Mastery Perpendicular Lines The perpendicular bisector of a segment is a line perpendicular to the segment at the segments midpoint. B 2 3 The distance from a point to a line is the length of the shortest segment from the point to the line. It is the length of the perpendicular segment that joins them. 3 4 7 X 5 You can write and solve an inequality for x. WU WT WT is the shortest segment. x 1 8 Substitute x 1 for WU and 8 for WT. 1 1 Subtract 1 from both sides of the equality. x 7 Use the figure for Exercises 1 and 2. 1. Name the shortest segment from point K to LN. 2. Write and solve an inequality for x. - X. Use the figure for Exercises 3 and 4. 3. Name the shortest segment from point Q to GH. 4. Write and solve an inequality for x. ( 1 X Line b is the perpendicular bisector of RS. The shortest segment from W to SU is WT. Copyright by Holt, Rinehart and Winston. 36 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Perpendicular Lines continued 3-4 You can use the following theorems about perpendicular lines in your proofs. Theorem Example If two intersecting lines form a linear pair of congruent angles, then the lines are perpendicular. Symbols: 2 intersecting lines form lin. pair of lines . A B 1 2 1 and 2 form a linear pair and 1 2, so a b. Perpendicular Transversal Theorem In a plane, if a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other line. Symbols: Transv. Thm. D C H h c and c d, so h d. If two coplanar lines are perpendicular to the same line, then the two lines are parallel to each other. Symbols: 2 lines to same line 2 lines . K J j and k , so j k. 5. Complete the two-column proof. Given: 1 2, s t Prove: r t Proof: Statements Reasons 1. 1 2 1. Given 2. a. 2. Conv. of Alt. Int. Thm. 3. s t 3. b. 4. r t 4. c. T S 1 2 R Copyright by Holt, Rinehart and Winston. 37 Holt Geometry All rights reserved. Name Date Class LESSON 3-5 Review for Mastery Slopes of Lines The slope of a line describes how steep the line is. You can find the slope by writing the ratio of the rise to the run. slope rise run 3 6 1 2 You can use a formula to calculate the slope m of the line through points (x 1. y 1 ) and (x 2. y 2 ). m rise run y 2 y 1 x 2 x 1 To find the slope of AB using the formula, substitute (1, 3) for (x 1. y 1 ) and (7, 6) for (x 2. y 2 ). Use the slope formula to determine the slope of each line. 0 X Y 2 2 2 ( 2 0 X Y 2 2 1. HJ 2. CD 0 X Y 2. - 2 3 0 X Y 2 2 3 2 2 3. LM 4. RS Change in x-values 0 X Y 4 (7, 6) (1, 3) 4 rise: go up 3 units run: go right 6 units m y 2 y 1 x 2 x 1 Slope formula 6 3 7 1 Substitution 3 6 Simplify. 1 2 Simplify. Change in y-values Copyright by Holt, Rinehart and Winston. 38 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Slopes of Lines continued 3-5 Slopes of Parallel and Perpendicular Lines 0 X Y 2. -. 0 2 4 2 slope of LM 3 slope of NP 3 Parallel lines have the same slope. 0 X Y 2. 0 1 2 2 2 4 slope of NP 3 slope of QR 1 3 product of slopes: 3 1 3 1 Perpendicular lines have slopes that are opposite reciprocals. The product of the slopes is 1. Use slopes to determine whether each pair of distinct lines is parallel, perpendicular, or neither. 5. slope of PQ 5 6. slope of EF 3 4 slope of JK 1 5 slope of CD 3 4 7. slope of BC 5 3 8. slope of WX 1 2 slope of ST 3 5 slope of YZ 1 2 Graph each pair of lines. Use slopes to determine whether the lines are parallel, perpendicular, or neither. X Y X Y 9. FG and HJ for F(1, 2), G(3, 4), 10. RS and TU for R(2, 3), S(3, 3), H(2, 3), and J(4, 1) T(3, 1), and U(3, 1) Copyright by Holt, Rinehart and Winston. 39 Holt Geometry All rights reserved. Name Date Class LESSON slope y-intercept slope Review for Mastery Lines in the Coordinate Plane 3-6 Slope-Intercept Form Point-Slope Form y mx b y 4x 7 y y 1 m(x x 1 ) point on the line: y 2 1 3 (x 5) (x 1. y 1 ) (5, 2) Write the equation of the line through (0, 1) and (2, 7) in slope-intercept form. Step 1: Find the slope. m y 2 y 1 x 2 x 1 Formula for slope 7 1 2 0 6 2 3 Step 2: Find the y-intercept. y mx b Slope-intercept form 1 3(0) b Substitute 3 for m, 0 for x, and 1 for y. 1 b Simplify. Step 3: Write the equation. y mx b Slope-intercept form y 3x 1 Substitute 3 for m and 1 for b. Write the equation of each line in the given form. 1. the line through (4, 2) and (8, 5) in 2. the line through (4, 6) with slope 1 2 slope-intercept form in point-slope form 3. the line through (5, 1) with slope 2 4. the line with x-intercept 5 and in point-slope form y-intercept 3 in slope-intercept form 5. the line through (8, 0) with slope 3 4 6. the line through (1, 7) and (6, 7) in slope-intercept form in point-slope form Copyright by Holt, Rinehart and Winston. 40 Holt Geometry All rights reserved. Name Date Class LESSON 3-6 Review for Mastery Lines in the Coordinate Plane continued You can graph a line from its equation. Consider the equation y 2 3 x 2. y-intercept 2 slope 2 3 X Y First plot the y-intercept (0, 2). Use rise 2 and run 3 to find another point. Draw the line containing the two points. Parallel Lines Intersecting Lines Coinciding Lines X Y same slope different y-intercepts X Y different slopes X Y same slope same y-intercept Graph each line. X Y X Y X Y 7. y x 2 8. y 1 3 x 3 9. y 2 1 4 (x 1) Determine whether the lines are parallel, intersect, or coincide. 10. y 2x 5 11. y 1 3 x 4 y 2x 1 x 3y 12 12. y 5x 2 13. 5y 2x 1 x 4y 8 y 2 5 x 3 run: go left 3 units rise: go up 2 units y 1 3 x 2 y 1 3 x y 1 2 x 2 y 2x 1 y 2 3 x 1 2x 3y 3 Copyright by Holt, Rinehart and Winston. 41 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Classifying Triangles 4-1 You can classify triangles by their angle measures. An equiangular triangle, for example, is a triangle with three congruent angles. Examples of three other triangle classifications are shown in the table. Acute Triangle Right Triangle Obtuse Triangle all acute angles one right angle one obtuse angle You can use angle measures to classify JML at right. JLM and JLK form a linear pair, so they are supplementary. mJLM mJLK 180 Def. of supp. mJLM 120 180 Substitution mJLM 60 Subtract. Since all the angles in JLM are congruent, JLM is an equiangular triangle. Classify each triangle by its angle measures. 1. 2. 3. Use the figure to classify each triangle by its angle measures. 4. DFG 5. DEG 6. EFG. N is equiangular. 60 60 60 - . JKL is obtuse so JLK is an obtuse triangle. Copyright by Holt, Rinehart and Winston. 42 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Classifying Triangles continued 4-1 You can also classify triangles by their side lengths. Equilateral Triangle Isosceles Triangle Scalene Triangle all sides congruent at least two sides congruent no sides congruent You can use triangle classification to find the side lengths of a triangle. Step 1 Find the value of x. QR RS Def. of segs. 4x 3x 5 Substitution x 5 Simplify. Step 2 Use substitution to find the length of a side. 4x 4(5) Substitute 5 for x. 20 Simplify. Each side length of QRS is 20. Classify each triangle by its side lengths. 7. EGF 8. DEF 9. DFG Find the side lengths of each triangle. 10. X X 11. X X X X X 2 1 3 Copyright by Holt, Rinehart and Winston. 43 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Angle Relationships in Triangles 4-2 mC 90 39 51 According to the Triangle Sum Theorem, the sum of the angle . measures of a triangle is 180. mJ mK mL 62 73 45 180 The corollary below follows directly from the Triangle Sum Theorem. Corollary Example The acute angles of a right triangle are complementary. mC mE 90 Use the figure for Exercises 1 and 2. 1. Find mABC. . 2. Find mCAD. Use RST for Exercises 3 and 4. 3. What is the value of x (7X 13) (4X 9) (2X 2) 2 4 3 4. What is the measure of each angle What is the measure of each angle . Aufrechtzuerhalten. X 5 7 6 5. L 6. C 7. W Copyright by Holt, Rinehart and Winston. 44 Holt Geometry All rights reserved. Name Date Class LESSON An exterior angle of a triangle is formed by one side of the triangle and the extension of an adjacent side. 1 and 2 are the remote interior angles of 4 because they are not adjacent to 4. Exterior Angle Theorem The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles. Third Angles Theorem If two angles of one triangle are congruent to two angles of another triangle, then the third pair of angles are congruent. Find each angle measure. ( 68 (4X 5) 3X. 8. mG 9. mD Find each angle measure. (6X 10) (7X 2). . 0 1 - X X 3 5 4 2 10. mM and mQ 11. mT and mR 4-2 Review for Mastery Angle Relationships in Triangles continued remote interior angles exterior angle m4 m1 m2 Copyright by Holt, Rinehart and Winston. 45 Holt Geometry All rights reserved. Name Date Class LESSON Triangles are congruent if they have the same size and shape. Their corresponding parts, the angles and sides that are in the same positions, are congruent. NN. Corresponding Parts Congruent Angles Congruent Sides A J B K C L AB JK BC KL CA LJ To identify corresponding parts of congruent triangles, look at the order of the vertices in the congruence statement such as ABC JKL. Given: XYZ NPQ. Identify the congruent corresponding parts. 9 1 8. 0 1. Z 2. YZ 3. P 4. X 5. NQ 6. PN Given: EFG RST. Find each value below. 2 3 4 (4X 6) (5Y 2) 3Z 8 Z 4 28 7. x 8. y 9. mF 10. ST 4-3 Review for Mastery Congruent Triangles Copyright by Holt, Rinehart and Winston. 46 Holt Geometry All rights reserved. Name Date Class LESSON You can prove triangles congruent by using the definition of congruence. Given: D and B are right angles. . DCE BCA C is the midpoint of DB. ED AB. EC AC Prove: EDC ABC Proof: Statements Reasons 1. D and B are rt. . 1. Given 2. D B 2. Rt. Thm. 3. DCE BCA 3. Given 4. E A 4. Third Thm. 5. C is the midpoint of DB. 5. Given 6. DC BC 6. Def. of mdpt. 7. ED AB. EC AC 7. Given 8. EDC ABC 8. Def. of s 11. Complete the proof. Given: Q R. 2 0 1 3 P is the midpoint of QR. NQ SR. NP SP Prove: NPQ SPR Proof: Statements Reasons 1. Q R 1. Given 2. NPQ SPR 2. a. 3. N S 3. b. 4. P is the midpoint of QR. 4. c. 5. d. 5. Def. of mdpt. 6. NQ SR. NP SP 6. e. 7. NPQ SPR 7. f. Review for Mastery Congruent Triangles continued 4-3 Copyright by Holt, Rinehart and Winston. 47 Holt Geometry All rights reserved. Name Date Class LESSON Side-Side-Side (SSS) Congruence Postulate If three sides of one triangle are congruent to three sides 0 5 4 2 3 1 of another triangle, then the triangles are congruent. QR TU. RP US. and PQ ST. so PQR STU. You can use SSS to explain why FJH FGH. ( It is given that FJ FG and that JH GH. By the Reflex. Prop. of , FH FH. So FJH FGH by SSS. Side-Angle-Side (SAS) Congruence Postulate If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. (. . - N(N,-. Use SSS to explain why the triangles in each pair are congruent. - . 1. JKM LKM 2. ABC CDA 3. Use SAS to explain why WXY WZY. 7 9. 8 4-4 Review for Mastery Triangle Congruence: SSS and SAS K is the included angle of HK and KJ. N is the included angle of LN and NM. Copyright by Holt, Rinehart and Winston. 48 Holt Geometry All rights reserved. Name Date Class LESSON You can show that two triangles are congruent by using SSS and SAS. Show that JKL FGH for y 7. HG y 6 mG 5y 5 FG 4y 1 7 6 13 5(7) 5 40 4(7) 1 27 HG LK 13, so HG LK by def. of segs. mG 40, so G K by def. of . FG JK 27, so FG JK by def. of segs. Therefore JKL FGH by SAS. Show that the triangles are congruent for the given value of the variable. ( X 2 X 2X 3 9 6 8 6 8 7 1 2 0 3N 8 7N 17 21 (36N 5) 113 4. BCD FGH, x 6 5. PQR VWX, n 3 6. Complete the proof. Given: T is the midpoint of VS. 6 3 2 4 RT VS Prove: RST RVT Statements Reasons 1. T is the midpoint of VS. 1. Given 2. a. 2. Def. of mdpt. 3. RT VS 3. b. 4. 4. c. 5. d. 5. Rt. Thm. 6. RT RT 6. e. 7. RST RVT 7. f. 4-4 Review for Mastery Triangle Congruence: SSS and SAS continued (. 27 13 4Y 1 (5Y 5) Y 6 40 Copyright by Holt, Rinehart and Winston. 49 Holt Geometry All rights reserved. Name Date Class LESSON Angle-Side-Angle (ASA) Congruence Postulate If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. NN Determine whether you can use ASA to prove the triangles congruent. Explain. . 0 1 CM CM - 8. M 9 M 1. KLM and NPQ 2. EFG and XYZ. - 0 7 5 6 4 3 3. KLM and PNM, given that M is the 4. STW and UTV midpoint of NL Review for Mastery Triangle Congruence: ASA, AAS, and HL 4-5 AC is the included side of A and C. DF is the included side of D and F. Copyright by Holt, Rinehart and Winston. 50 Holt Geometry All rights reserved. Name Date Class LESSON Angle-Angle-Side (AAS) Congruence Theorem If two angles and a nonincluded side of one triangle are congruent to the corresponding angles and nonincluded side of another triangle, then the triangles are congruent. . ( N(N, Special theorems can be used to prove right triangles congruent. Hypotenuse-Leg (HL) Congruence Theorem If the hypotenuse and a leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. . 0 - N, N-.0 5. Describe the corresponding parts and the justifications . for using them to prove the triangles congruent by AAS. Given: BD is the angle bisector of ADC. Prove: ABD CBD Determine whether you can use the HL Congruence Theorem to prove the triangles congruent. If yes, explain. If not, tell what else you need to know. 5 8 7 6 2 4 3 1 0 6. UVW WXU 7. TSR PQR 4-5 Review for Mastery Triangle Congruence: ASA, AAS, and HL continued FH is a nonincluded side of F and G. JL is a nonincluded side of J and K. Copyright by Holt, Rinehart and Winston. 51 Holt Geometry All rights reserved. Name Date Class LESSON Corresponding Parts of Congruent Triangles are Congruent (CPCTC) is useful in proofs. If you prove that two triangles are congruent, then you can use CPCTC as a justification for proving corresponding parts congruent. Given: AD CD. AB CB Prove: A C Proof. Given SSS CPCTC NN Reflex. Prop of Given Complete each proof. 0 1. -. 1. Given: PNQ LNM, PN LN. N is the midpoint of QM. Prove: PQ LM Proof: 0. SAS 01,- Given C D A 0.1,.- Given. is the mdpt. of -1. Def. of midpt. B 2. Given: UXW and UVW are right s. 5 6 8 7 UX UV Prove: X V Proof: Statements Reasons 1. UXW and UVW are rt. S 1. Given 2. UX UV 2. a. 3. UW UW 3. b. 4. c. 4. d. 5. X V 5. e. Review for Mastery Triangle Congruence: CPCTC 4-6 . Copyright by Holt, Rinehart and Winston. 52 Holt Geometry All rights reserved. Name Date Class LESSON You can also use CPCTC when triangles are on the coordinate plane. Given: C(2, 2), D(4, 2), E(0, 2), 0 X ( Y 2 2 2 F(0, 1), G(4, 1), H(4, 3) Prove: CED FHG Step 1 Plot the points on a coordinate plane. Step 2 Find the lengths of the sides of each triangle. Use the Distance Formula if necessary. d (x 2 x 1 ) 2 (y 2 y 1 ) 2 CD (4 2) 2 (2 2) 2 FG (4 0) 2 (1 1) 2 4 16 2 5 16 4 2 5 DE 4 GH 4 EC (2 0) 2 2 (2) 2 HF 0 (4) 2 (1 3) 2 4 16 2 5 16 4 2 5 So, CD FG. DE GH. and EC HF. Therefore CDE FGH by SSS, and CED FHG by CPCTC. Use the graph to prove each congruence statement. 0 X 9 8 7 3 1 2 Y 2 2 3 2 0 X . Y 2 3 2 2 3. RSQ XYW 4. CAB LJK 5. Use the given set of points to prove PMN VTU. M(2, 4), N(1, 2), P(3, 4), T(4, 1), U(2, 4), V(4, 0) 4-6 Review for Mastery Triangle Congruence: CPCTC continued Copyright by Holt, Rinehart and Winston. 53 Holt Geometry All rights reserved. Name Date Class LESSON A coordinate proof is a proof that uses coordinate geometry and algebra. In a coordinate proof, the first step is to position a figure in a plane. There are several ways you can do this to make your proof easier. Positioning a Figure in the Coordinate Plane Keep the figure in 0 X Y 2 2 Quadrant I by using the origin as a vertex. Center the figure 0 X Y 2 2 3 3 at the origin. Center a side of the 0 X Y 3 3 3 figure at the origin. Use one or both axes 0 X Y 3 3 as sides of the figure. Position each figure in the coordinate plane and give the coordinates of each vertex. X Y X Y 1. a square with side lengths of 6 units 2. a right triangle with leg lengths of 3 units and 4 units X Y X Y 3. a triangle with a base of 8 units and 4. a rectangle with a length of 6 units and a height of 2 units a width of 3 units 4-7 Review for Mastery Introduction to Coordinate Proof Copyright by Holt, Rinehart and Winston. 54 Holt Geometry All rights reserved. Name Date Class LESSON You can prove that a statement about a figure is true without knowing the side lengths. To do this, assign variables as the coordinates of the vertices. X Y D C Position each figure in the coordinate plane and give the coordinates of each vertex. 5. a right triangle with leg lengths s and t 6. a square with side lengths k 7. a rectangle with leg lengths and w 8. a triangle with base b and height h 9. Describe how you could use the formulas for midpoint and slope to prove the following. Given: HJK, R is the midpoint of HJ. S is the midpoint of JK. Prove: RS HK 4-7 Review for Mastery Introduction to Coordinate Proof continued a right triangle with leg lengths c and d Copyright by Holt, Rinehart and Winston. 55 Holt Geometry All rights reserved. Name Date Class LESSON Theorem Examples Isosceles Triangle Theorem If two sides of a triangle are congruent, then the angles opposite the sides are congruent. 4 3 2 If RT RS. then T S. Converse of Isosceles Triangle Theorem If two angles of a triangle are congruent, then the sides opposite those angles are congruent. Aufrechtzuerhalten. If N M, then LN LM. You can use these theorems to find angle measures in isosceles triangles. Find mE in DEF. mD mE Isosc. Thm. X X 5x (3x 14) Substitute the given values. 2x 14 Subtract 3x from both sides. x 7 Divide both sides by 2. Thus mE 3(7) 14 35. Find each angle measure. 2 0 1 1. mC 2. mQ ( X X. -. X X 3. mH 4. mM Review for Mastery Isosceles and Equilateral Triangles 4-8 Copyright by Holt, Rinehart and Winston. 56 Holt Geometry All rights reserved. Name Date Class LESSON 4-8 Equilateral Triangle Corollary If a triangle is equilateral, then it is equiangular. (equilateral equiangular ) Equiangular Triangle Corollary If a triangle is equiangular, then it is equilateral. (equiangular equilateral ) If A B C, then AB BC CA. You can use these theorems to find values in equilateral triangles. Find x in STV. STV is equiangular. Equilateral equiangular 3 4 6 X (7x 4) 60 The measure of each of an equiangular is 60. 7x 56 Subtract 4 from both sides. x 8 Divide both sides by 7. Find each value. 1 2 3 N X 5. n 6. x 3 4 6 R R -. Y Y 7. VT 8. MN Review for Mastery Isosceles and Equilateral Triangles continued. Copyright by Holt, Rinehart and Winston. 57 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Perpendicular and Angle Bisectors 5-1 Theorem Example Perpendicular Bisector Theorem If a point is on the perpendicular bisector of a segment, then it is equidistant, or the same distance, from the endpoints of the segment. Given: is the perpendicular bisector of FG. Conclusion: AF AG The Converse of the Perpendicular Bisector Theorem is also true. If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment. You can write an equation for the perpendicular bisector of a segment. Consider the segment with endpoints Q (5, 6) and R (1, 2). Step 1 Find the midpoint of QR. Step 2 Find the slope of the bisector of QR. x 1 x 2 2. y 1 y 2 2 5 1 2. 6 2 2 y 2 y 1 x 2 x 1 2 6 1 (5) Slope of QR (2, 4) 2 3 So the slope of the bisector of QR is 3 2. Step 3 Use the point-slope form to write an equation. y y 1 m (x x 1 ) Point-slope form y 4 3 2 (x 2) Slope 3 2 line passes through (2, 4), the midpoint of QR. Find each measure. 2 M 4 3 16 14 6 T . 2.5 4 3X 1 5X 3 (. 1. RT 2. AB 3. HJ Write an equation in point-slope form for the perpendicular bisector of the segment with the given endpoints. 4. A (6, 3), B (0, 5) 5. W (2, 7), X (4, 3) Each point on is equidistant from points F and G. Copyright by Holt, Rinehart and Winston. 58 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Perpendicular and Angle Bisectors continued 5-1 Theorem Example Angle Bisector Theorem If a point is on the bisector of an angle, then it is equidistant from the sides of the angle. - 0. Given: MP is the angle bisector of LMN. Conclusion: LP NP Converse of the Angle Bisector Theorem If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle. - 0. Given: LP NP Conclusion: MP is the angle bisector of LMN. Find each measure. ( 2 4 3 1 8. 7 9 X X 6. EH 7. mQRS 8. mWXZ Use the figure for Exercises 911. ( . 9. Given that JL bisects HJK and LK 11.4, find LH. 10. Given that LH 26, LK 26, and mHJK 122, find mLJK. 11. Given that LH LK, mHJL (3y 19), and mLJK (4y 5), find the value of y. Point P is equidistant from sides ML and MN. LMP NMP Copyright by Holt, Rinehart and Winston. 59 Holt Geometry All rights reserved. Name Date Class LESSON 4. 1 3 - 2 0 Theorem Example Circumcenter Theorem The circumcenter of a triangle is equidistant from the vertices of the triangle. Given: MR. MS. and MT are 4. 1 3 - 2 0 the perpendicular bisectors of NPQ. Conclusion: MN MP MQ If a triangle on a coordinate plane has two sides that lie along the axes, you can easily find the circumcenter. Find the equations for the perpendicular bisectors of those two sides. The intersection of their graphs is the circumcenter. HD. JD. and KD are the perpendicular bisectors of EFG. ( Find each length. 1. DG 2. EK 3. FJ 4. DE Find the circumcenter of each triangle. 5. X Y / 6. X Y ,(0, 5) -(8, 0) /(0, 0) 3 4 Review for Mastery Bisectors of Triangles 5-2 The point of intersection of MR. MS. and MT is called the circumcenter of NPQ. Perpendicular bisectors MR. MS. and MT are concurrent because they intersect at one point. Copyright by Holt, Rinehart and Winston. 60 Holt Geometry All rights reserved. Name Date Class LESSON (. Theorem Example Incenter Theorem The incenter of a triangle is equidistant from the sides of the triangle. Given: AG. AH. and AJ are (. the angle bisectors of GHJ. Conclusion: AB AC AD WM and WP are angle bisectors of MNP, and WK 21. - 0. 7 Find mWPN and the distance from W to MN and NP. mNMP 2mNMW Def. of bisector mNMP 2(32) 64 Substitute. mNMP mN mNPM 180 Sum Thm. 64 72 mNPM 180 Substitute. mNPM 44 Subtract 136 from each side. mWPN 1 2 mNPM Def. of bisector mWPN 1 2 (44) 22 Substitute. The distance from W to MN and NP is 21 by the Incenter Theorem. PC and PD are angle bisectors of CDE. Find each measure. 1 0 7. the distance from P to CE 8. mPDE KX and KZ are angle bisectors of XYZ. Find each measure. 8 9 9. the distance from K to YZ 10. mKZY 5-2 Review for Mastery Bisectors of Triangles continued The point of intersection of AG. AH. and AJ is called the incenter of GHJ. Angle bisectors of GHJ intersect at one point. Copyright by Holt, Rinehart and Winston. 61 Holt Geometry All rights reserved. Name Date Class LESSON . (. Theorem Example Centroid Theorem The centroid of a triangle is located 2 3 of the distance from each vertex to the midpoint of the opposite side. . (. Given: AH. CG. and BJ are medians of ABC. Conclusion: AN 2 3 AH, CN 2 3 CG, BN 2 3 BJ In ABC above, suppose AH 18 and BN 10. You can use the Centroid Theorem to find AN and BJ. AN 2 3 AH Centroid Thm. BN 2 3 BJ Centroid Thm. AN 2 3 (18) Substitute 18 for AH. 10 2 3 BJ Substitute 10 for BN. AN 12 Simplify. 15 BJ Simplify. In QRS, RX 48 and QW 30. Find each length. 8 1 3. 9 7 2 1. RW 2. WX 3. QZ 4. WZ In HJK, HD 21 and BK 18. Find each length. ( 5. HB 6. BD 7. CK 8. CB 5-3 Review for Mastery Medians and Altitudes of Triangles The point of intersection of the medians is called the centroid of ABC. AH. BJ. and CG are medians of a triangle. They each join a vertex and the midpoint of the opposite side. Copyright by Holt, Rinehart and Winston. 62 Holt Geometry All rights reserved. Name Date Class LESSON . Find the orthocenter of ABC with vertices A (3, 3), B (3, 7), and C (3, 0). Step 1 Graph the triangle. X Y . Step 2 Find equations of the lines containing two altitudes. The altitude from A to BC is the horizontal line y 3. The slope of AC 0 3 3 (3) 1 2. so the slope of the altitude from B to AC is 2. The altitude must pass through B(3, 7). y y 1 m(x x 1 ) Point-slope form y 7 2(x 3) Substitute 2 for m and the coordinates of B (3, 7) for (x 1. y 1 ). y 2x 1 Simplify. Step 3 Solving the system of equations y 3 and y 2x 1, you find that the coordinates of the orthocenter are (1, 3). Triangle FGH has coordinates F (3, 1), G (2, 6), and H (4, 1). 9. Find an equation of the line containing the X Y ( altitude from G to FH. 10. Find an equation of the line containing the altitude from H to FG. 11. Solve the system of equations from Exercises 9 and 10 to find the coordinates of the orthocenter. Find the orthocenter of the triangle with the given vertices. 12. N (1, 0), P (1, 8), Q (5, 0) 13. R (1, 4), S (5, 2), T (1, 6) Review for Mastery Medians and Altitudes of Triangles continued 5-3 The point of intersection of the altitudes is called the orthocenter of JKL. JD. KE. and LC are altitudes of a triangle. They are perpendicular segments that join a vertex and the line containing the side opposite the vertex. Copyright by Holt, Rinehart and Winston. 63 Holt Geometry All rights reserved. Name Date Class LESSON A midsegment of a triangle joins the midpoints of two sides of the triangle. Every triangle has three midsegments. 3 2 Use the figure for Exercises 14. AB is a midsegment of RST. 1. What is the slope of midsegment AB and the slope X Y 3(2, 3) (1, 0) 4(6, 1) (3, 2) 2(0, 3) 2 2 3 0 of side ST. 2. What can you conclude about AB and ST. 3. Find AB and ST. 4. Compare the lengths of AB and ST. Use MNP for Exercises 57. 5. UV is a midsegment of MNP. Find the X Y .(4, 5) 5 0(2, 1) 6 -(4, 7) 4 3 3 0 coordinates of U and V. 6. Show that UV MN. 7. Show that UV 1 2 MN. 5-4 Review for Mastery The Triangle Midsegment Theorem RS is a midsegment of CDE. R is the midpoint of CD. S is the midpoint of CE. Copyright by Holt, Rinehart and Winston. 64 Holt Geometry All rights reserved. Name Date Class LESSON Theorem Example Triangle Midsegment Theorem A midsegment of a triangle is parallel to a side of the triangle, and its length is half the length of that side. 1. 0 - Given: PQ is a midsegment of LMN. Conclusion: PQ LN. PQ 1 2 LN You can use the Triangle Midsegment Theorem to (. find various measures in ABC. HJ 1 2 AC Midsegment Thm. HJ 1 2 (12) Substitute 12 for AC. HJ 6 Simplify. JK 1 2 AB Midsegment Thm. HJ AC Midsegment Thm. 4 1 2 AB Substitute 4 for JK. mBCA mBJH Corr. Thm. 8 AB Simplify. mBCA 35 Substitute 35 for mBJH. Find each measure. ( 6 8 7 8. VX 9. HJ 10. mVXJ 11. XJ Find each measure. 3 4 2 12. ST 13. DE 14. mDES 15. mRCD 5-4 Review for Mastery The Triangle Midsegment Theorem continued Copyright by Holt, Rinehart and Winston. 65 Holt Geometry All rights reserved. Name Date Class LESSON In a direct proof, you begin with a true hypothesis and prove that a conclusion is true. In an indirect proof, you begin by assuming that the conclusion is false (that is, that the opposite of the conclusion is true). You then show that this assumption leads to a contradiction. Consider the statement Two acute angles do not form a linear pair. Writing an Indirect Proof Steps Example 1. Identify the conjecture to be proven. Given: 1 and 2 are acute angles. Prove: 1 and 2 do not form a linear pair. 2. Assume the opposite of the conclusion is true. Assume 1 and 2 form a linear pair. 3. Use direct reasoning to show that the assumption leads to a contradiction. m1 m2 180 by def. of linear pair. Since m1 90 and m2 90, m1 m2 180. This is a contradiction. 4. Conclude that the assumption is false and hence that the original conjecture must be true. The assumption that 1 and 2 form a linear pair is false. Therefore 1 and 2 do not form a linear pair. Use the following statement for Exercises 14. . An obtuse triangle cannot have a right angle. 1. Identify the conjecture to be proven. 2. Assume the opposite of the conclusion. Write this assumption. 3. Use direct reasoning to arrive at a contradiction. 4. What can you conclude Review for Mastery Indirect Proof and Inequalities in One Triangle 5-5 Copyright by Holt, Rinehart and Winston. 66 Holt Geometry All rights reserved. Name Date Class LESSON Theorem Example If two sides of a triangle are not congruent, then the larger angle is opposite the longer side. 8 7 9 If WY XY, then mX mW. Another similar theorem says that if two angles of a triangle are not congruent, then the longer side is opposite the larger angle. Write the correct answer. 6 3 4 ( 5. Write the angles in order from smallest 6. Write the sides in order from shortest to largest. to longest. Theorem Example Triangle Inequality Theorem The sum of any two side lengths of a triangle is greater than the third side length. A C B a b c b c a c a b Tell whether a triangle can have sides with the given lengths. Explain. 7. 3, 5, 8 8. 11, 15, 21 5-5 Review for Mastery Indirect Proof and Inequalities in One Triangle continued WY is the longest side. X is the largest angle. Copyright by Holt, Rinehart and Winston. 67 Holt Geometry All rights reserved. Name Date Class LESSON Theorem Example Hinge Theorem If two sides of one triangle are congruent to two sides of another triangle and the included angles are not congruent, then the included angle that is larger has the longer third side across from it. - (. If K is larger than G, then side LM is longer than side HJ. The Converse of the Hinge Theorem is also true. In the example above, if side LM is longer than side HJ. then you can conclude that K is larger than G. You can use both of these theorems to compare various measures of triangles. Compare NR and PQ in the figure at right. 0 1 2 3 PN QR PR PR mNPR mQRP Since two sides are congruent and NPR is smaller than QRP, the side across from it is shorter than the side across from QRP. So NR PQ by the Hinge Theorem. Compare the given measures. 3 8 9 7 6 4 ( -. 1. TV and XY 2. mG and mL . ( 3. AB and AD 4. mFHE and mHFG Review for Mastery Inequalities in Two Triangles 5-6 Copyright by Holt, Rinehart and Winston. 68 Holt Geometry All rights reserved. Name Date Class LESSON You can use the Hinge Theorem and its converse to find a range of values in triangles. Use MNP and QRS to find the range of values for x. 1 2 3 0 - X Step 1 Compare the side lengths in the triangles. NM SR NP SQ mN mS Since two sides of MNP are congruent to two sides of QRS and mN mS, then MP QR by the Hinge Theorem. MP QR Hinge Thm. 3x 6 24 Substitute the given values. 3x 30 Add 6 to each side. x 10 Divide each side by 3. Step 2 Check that the measures are possible for a triangle. Since MP is in a triangle, its length must be greater than 0. MP 0 Def. of 3x 6 0 Substitute 3x 6 for MP. x 2 Simplify. Step 3 Combine the inequalities. A range of values for x is 2 x 10. Find a range of values for x. 5. X 6. 26 23 (3X 9) 54 7. 14 10 (2X 6) 108 8. X 5-6 Review for Mastery Inequalities in Two Triangles continued Copyright by Holt, Rinehart and Winston. 69 Holt Geometry All rights reserved. Name Date Class LESSON The Pythagorean Theorem states that the following relationship exists among the lengths of the legs, a and b, and the length of the hypotenuse, c, of any right triangle. A C B a 2 b 2 c 2 Use the Pythagorean Theorem to find the value of x in each triangle. X X X a 2 b 2 c 2 Pythagorean Theorem a 2 b 2 c 2 x 2 6 2 9 2 Substitute. x 2 4 2 (x 2) 2 x 2 36 81 Take the squares. x 2 16 x 2 4x 4 x 2 45 Simplify. 4x 12 x 45 x 3 x 3 5 Find the value of x. Give your answer in simplest radical form. 1. X 2. X 3. X 4. X X 5-7 Review for Mastery The Pythagorean Theorem Take the positive square root and simplify. Copyright by Holt, Rinehart and Winston. 70 Holt Geometry All rights reserved. Name Date Class LESSON A Pythagorean triple is a set of three nonzero whole numbers a, b, and c that satisfy the equation a 2 b 2 c 2. You can use the following theorem to classify triangles by their angles if you know their side lengths. Always use the length of the longest side for c. Pythagorean Inequalities Theorem. C A B If c 2 a 2 b 2. then ABC is obtuse. C A B If c 2 a 2 b 2. then ABC is acute. Consider the measures 2, 5, and 6. They can be the side lengths of a triangle since 2 5 6, 2 6 5, and 5 6 2. If you substitute the values into c 2 a 2 b 2. you get 36 29. Since c 2 a 2 b 2. a triangle with side lengths 2, 5, and 6 must be obtuse. Find the missing side length. Tell whether the side lengths form a Pythagorean triple. Erklären. 5. 6. Tell whether the measures can be the side lengths of a triangle. If so, classify the triangle as acute, obtuse, or right. 7. 4, 7, 9 8. 10, 13, 16 9. 8, 8, 11 10. 9, 12, 15 11. 5, 14, 20 12. 4.5, 6, 10.2 5-7 Review for Mastery The Pythagorean Theorem continued Pythagorean Triples Not Pythagorean Triples 3, 4, 5, 5, 12, 13 2, 3, 4 6, 9, 117 mC 90 mC 90 Copyright by Holt, Rinehart and Winston. 71 Holt Geometry All rights reserved. Name Date Class LESSON Theorem Example 45-45-90 Triangle Theorem In a 45-45-90 triangle, both legs are congruent and the length of the hypotenuse is 2 times the length of a leg. qi qi In a 45-45-90 triangle, if a leg X X Xqi length is x, then the hypotenuse length is x 2. Use the 45-45-90 Triangle Theorem to find the value of x in EFG. Every isosceles right triangle is a 45-45-90 triangle. Triangle X X EFG is a 45-45-90 triangle with a hypotenuse of length 10. 10 x 2 Hypotenuse is 2 times the length of a leg. 10 2 x 2 2 Divide both sides by 2. 5 2 x Rationalize the denominator. Find the value of x. Give your answers in simplest radical form. 1. X 2. X 3. X X 4. X qi Review for Mastery Applying Special Right Triangles 5-8 Copyright by Holt, Rinehart and Winston. 72 Holt Geometry All rights reserved. Name Date Class LESSON 5-8 Theorem Examples 30-60-90 Triangle Theorem In a 30-60-90 triangle, the length of the hypotenuse is 2 multiplied by the length of the shorter leg, and the longer leg is 3 multiplied by the length of the shorter leg. qi qi In a 30-60-90 triangle, if the shorter leg Xqi X X length is x, then the hypotenuse length is 2x and the longer leg length is x. Use the 30-60-90 Triangle Theorem to find the values X Y ( of x and y in HJK. 12 x 3 Longer leg shorter leg multiplied by 3. 12 3 x Divide both sides by 3. 4 3 x Rationalize the denominator. y 2x Hypotenuse 2 multiplied by shorter leg. y 2(4 3 ) Substitute 4 3 for x. y 8 3 Simplify. Find the values of x and y. Give your answers in simplest radical form. 5. X Y 6. X Y 7. X Y qi 8. X Y Review for Mastery Applying Special Right Triangles continued Copyright by Holt, Rinehart and Winston. 73 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Properties and Attributes of Polygons 6-1 The parts of a polygon are named on the quadrilateral below. You can name a polygon by the number of its sides. A regular polygon has all sides congruent and all angles congruent. A polygon is convex if all its diagonals lie in the interior of the polygon. A polygon is concave if all or part of at least one diagonal lies outside the polygon. Types of Polygons regular, convex irregular, convex irregular, concave Tell whether each figure is a polygon. If it is a polygon, name it by the number of sides. 1. 2. 3. Tell whether each polygon is regular or irregular. Then tell whether it is concave or convex. 4. 5. 6. Number of Sides Polygon 3 triangle 4 quadrilateral 5 pentagon 6 hexagon 7 heptagon 8 octagon 9 nonagon 10 decagon n n-gon diagonal vertex side Copyright by Holt, Rinehart and Winston. 74 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Properties and Attributes of Polygons continued 6-1 The Polygon Angle Sum Theorem states that the sum of the interior angle measures of a convex polygon with n sides is (n 2)180. Convex Polygon Number of Sides Sum of Interior Angle Measures: (n 2)180 quadrilateral 4 (4 2)180 360 hexagon 6 (6 2)180 720 decagon 10 (10 2)180 1440 If a polygon is a regular polygon, then you can divide the sum of the interior angle measures by the number of sides to find the measure of each interior angle. Regular Polygon Number of Sides Sum of Interior Angle Measures Measure of Each Interior Angle quadrilateral 4 360 360 4 90 hexagon 6 720 720 6 120 decagon 10 1440 1440 10 144 The Polygon External Angle Sum Theorem states that the sum of the exterior angle measures, one angle at each vertex, of a convex polygon is 360. 152 145 152 63 145 360 63 The measure of each exterior angle of a regular polygon with n exterior angles is 360 n. So the measure of each exterior angle of a regular decagon is 360 10 36. Find the sum of the interior angle measures of each convex polygon. 7. pentagon 8. octagon 9. nonagon Find the measure of each interior angle of each regular polygon. Round to the nearest tenth if necessary. 10. pentagon 11. heptagon 12. 15-gon Find the measure of each exterior angle of each regular polygon. 13. quadrilateral 14. octagon Copyright by Holt, Rinehart and Winston. 75 Holt Geometry All rights reserved. Name Date Class LESSON 6-2 Review for Mastery Properties of Parallelograms A parallelogram is a quadrilateral with two pairs of parallel sides. All parallelograms, such as FGHJ, have the following properties. ( ( Properties of Parallelograms FG HJ GH JF Opposite sides are congruent. F H G J Opposite angles are congruent. mF mG 180 mG mH 180 mH mJ 180 mJ mF 180 Consecutive angles are supplementary. FP HP GP JP The diagonals bisect each other. Find each measure. 1. AB 2. mD. CM CM. Find each measure in LMNP. 3. ML 4. LP 5. mLPM 6. LN -. 0 1 62 32 10 m 12 m 9 m 7. mMLN 8. QN ( ( ( ( 0 Copyright by Holt, Rinehart and Winston. 76 Holt Geometry All rights reserved. Name Date Class LESSON 6-2 Review for Mastery Properties of Parallelograms continued You can use properties of parallelograms to find measures. WXYZ is a parallelogram. Find mX. 7. 8 9 X X mW mX 180 If a quadrilateral is a , then cons. are supp. (7x 15) 4x 180 Substitute the given values. 11x 15 180 Combine like terms. 11x 165 Subtract 15 from both sides. x 15 Divide both sides by 11. mX (4x) 4(15) 60 If you know the coordinates of three vertices of a parallelogram, you can use slope to find the coordinates of the fourth vertex. Three vertices of RSTV are R(3, 1), S(1, 5), and T(3, 6). Find the coordinates of V. Since opposite sides must be parallel, the rise and the run from S to R must be the same as the rise and the run from T to V. From S to R, you go down 4 units and right 4 units. So, from T to V, go down 4 units and right 4 units. Vertex V is at V(7, 2). You can use the slope formula to verify that ST RV. X Y 3 2 6 4 3 3 4 4 CDEF is a parallelogram. Find each measure. 9. CD 10. EF 3Z 4W 8 5W 1 (9Z 12) 11. mF 12. mE The coordinates of three vertices of a parallelogram are given. Find the coordinates of the fourth vertex. 13. ABCD with A(0, 6), B(5, 8), C(5, 5) 14. KLMN with K(4, 7), L(3, 6), M(5, 3) Copyright by Holt, Rinehart and Winston. 77 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Conditions for Parallelograms 6-3 You can use the following conditions to determine whether a quadrilateral such as PQRS is a parallelogram. 0 3 1 2 Conditions for Parallelograms QR SP QR SP If one pair of opposite sides is and , then PQRS is a parallelogram. QR SP PQ RS If both pairs of opposite sides are , then PQRS is a parallelogram. P R Q S If both pairs of opposite angles are , then PQRS is a parallelogram. PT RT QT ST If the diagonals bisect each other, then PQRS is a parallelogram. A quadrilateral is also a parallelogram if one of the angles is supplementary to both of its consecutive angles. 65 115 180, so A is supplementary to B and D. Therefore, ABCD is a parallelogram. . Show that each quadrilateral is a parallelogram for the given values. Erklären. 1. Given: x 9 and y 4 2. Given: w 3 and z 31 2 3 4 1 X Y Y X 4W 2 (3Z 25) 2Z W 7 0 3 1 2 0 3 1 2 0 3 1 2 0 3 1 2 4 Copyright by Holt, Rinehart and Winston. 78 Holt Geometry All rights reserved. Name Date Class LESSON You can show that a quadrilateral is a parallelogram by using any of the conditions listed below. Conditions for Parallelograms Both pairs of opposite sides are parallel (definition). One pair of opposite sides is parallel and congruent. Both pairs of opposite sides are congruent. Both pairs of opposite angles are congruent. The diagonals bisect each other. One angle is supplementary to both its consecutive angles. (. - EFGH must be a parallelogram JKLM may not be a parallelogram because both pairs of opposite because none of the sets of conditions sides are congruent. for a parallelogram is met. Determine whether each quadrilateral must be a parallelogram. Justify your answer. 3. 4. 5. 6. Show that the quadrilateral with the given vertices is a parallelogram by using the given definition or theorem. 7. J(2, 2), K(3, 3), L(1, 5), M(2, 0) 8. N(5, 1), P(2, 7), Q(6, 9), R(9, 3) Both pairs of opposite sides are parallel. Both pairs of opposite sides are congruent. 6-3 Review for Mastery Conditions for Parallelograms continued Copyright by Holt, Rinehart and Winston. 79 Holt Geometry All rights reserved. Name Date Class LESSON 6-4 Review for Mastery Properties of Special Parallelograms A rectangle is a quadrilateral with four right angles. A rectangle has the following properties. Properties of Rectangles ( ( is a parallelogram. If a quadrilateral is a rectangle, then it is a parallelogram. ( ( If a parallelogram is a rectangle, then its diagonals are congruent. Since a rectangle is a parallelogram, a rectangle also has all the properties of parallelograms. A rhombus is a quadrilateral with four congruent sides. A rhombus has the following properties. Properties of Rhombuses 2 3 4 1 1234 is a parallelogram. If a quadrilateral is a rhombus, then it is a parallelogram. 2 3 4 1 13 24 If a parallelogram is a rhombus, then its diagonals are perpendicular. 2 3 4 1 213314 If a parallelogram is a rhombus, then each diagonal bisects a pair of opposite angles. Since a rhombus is a parallelogram, a rhombus also has all the properties of parallelograms. ABCD is a rectangle. Find each length. 1. BD 2. CD 3. AC 4. AE 12 in. 5 in. 6.5 in. . KLMN is a rhombus. Find each measure. 5. KL 6. mMNK. . - (2Y 5) 9Y 3X 4 X 20 Copyright by Holt, Rinehart and Winston. 80 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Properties of Special Parallelograms continued 6-4 A square is a quadrilateral with four right angles and four congruent sides. A square is a parallelogram, a rectangle, and a rhombus. 2ECTANGLE parallelogram with 4 right 2HOMBUS parallelogram with 4 sides 0ARALLELOGRAM opposite sides are and 3QUARE parallelogram with 4 sides and 4 right Show that the diagonals of square HJKL are congruent perpendicular bisectors of each other. Step 1 Show that HK JL. HK (6 0) 2 (4 2) 2 2 10 JL (4 2) 2 (0 6) 2 2 10 HK JL 2 10. so HK JL. Step 2 Show that HK JL. slope of HK 4 2 6 0 1 3 slope of JL 0 6 4 2 3 Since the product of the slopes is 1, HK JL. Step 3 Show that HK and JL bisect each other by comparing their midpoints. midpoint of HK (3, 3) midpoint of JL (3, 3) Since they have the same midpoint, HK and JL bisect each other. The vertices of square ABCD are A(1, 0), B(4, 5), C(1, 8), and D(4, 3). Show that each of the following is true. 7. The diagonals are congruent. 8. The diagonals are perpendicular bisectors of each other. X Y ((0, 2) (2, 6) (6, 4) ,(4, 0) 3 3 Copyright by Holt, Rinehart and Winston. 81 Holt Geometry All rights reserved. Name Date Class LESSON 6-5 Review for Mastery Conditions for Special Parallelograms You can use the following conditions to determine whether a parallelogram is a rectangle. -. If one angle is a right angle, then JKLM is a rectangle. JL KM If the diagonals are congruent, then JKLM is a rectangle. You can use the following conditions to determine whether a parallelogram is a rhombus. 5 4 7 6 If one pair of consecutive sides are congruent, then TUVW is a rhombus. 5 4 7 6 If the diagonals are perpendicular, then TUVW is a rhombus. 5 4 7 6 If one diagonal bisects a pair of opposite angles, then TUVW is a rhombus. Determine whether the conclusion is valid. If not, tell what additional information is needed to make it valid. 1. EFGH is a rectangle. 2. MPQR is a rhombus. ( 0 - 2 1 For Exercises 3 and 4, use the figure to determine whether the conclusion is valid. If not, tell what additional information is needed to make it valid. 3. Given: EF GH. HE FG. EG FH Conclusion: EFGH is a rectangle. 4. Given: mEFG 90 Conclusion: EFGH is a rectangle. ( -. Copyright by Holt, Rinehart and Winston. 82 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Conditions for Special Parallelograms continued 6-5 You can identify special parallelograms in the coordinate plane by examining their diagonals. If the Diagonals are. the Parallelogram is a congruent rectangle perpendicular rhombus congruent and perpendicular square Use the diagonals to determine whether parallelogram ABCD is a rectangle, rhombus, or square. Give all the names that apply. Step 1 Find AC and BD to determine whether ABCD is a rectangle. AC (6 1) 2 (5 1) 2 41 BD (6 1) 2 (1 5) 2 41 Since 41 41. the diagonals are congruent. So ABCD is a rectangle. Step 2 Find the slopes of AC and BD to determine whether ABCD is a rhombus. slope of AC 5 1 6 1 4 5 slope of BD 1 5 6 1 4 5 Since 4 5 4 5 1, the diagonals are not perpendicular. So ABCD is not a rhombus and cannot be a square. Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. 5. V(3, 0), W(6, 4), X(11, 4), Y(8, 0) 6. L(1, 2), M(3, 5), N(6, 3), P(4, 0) 7. H(1, 3), J(10, 6), K(12, 0), L(3, 3) 8. E(4, 3), F(1, 2), G(2, 1), H(5, 0) X Y (1, 1) (1, 5) (6, 5) (6, 1) 3 3 0 Copyright by Holt, Rinehart and Winston. 83 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Properties of Kites and Trapezoids 6-6 A kite is a quadrilateral with exactly two pairs of congruent ( consecutive sides. If a quadrilateral is a kite, such as FGHJ, then it has the following properties. Properties of Kites FH GJ The diagonals are perpendicular. G J Exactly one pair of opposite angles is congruent. A trapezoid is a quadrilateral with exactly one pair of parallel sides. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. BASE BASE LEG LEG Isosceles Trapezoid Theorems In an isosceles trapezoid, each pair of base angles is congruent. If a trapezoid has one pair of congruent base angles, then it is isosceles. A trapezoid is isosceles if and only if its diagonals are congruent. In kite ABCD, mBCD 98, and mADE 47. Find each measure. 1. mDAE 2. mBCE 3. mABC . 4. Find mJ in trapezoid JKLM. 5. In trapezoid EFGH, FH 9. Find AG. . - 124 (. 6.2 ( ( Each side is called a base. Each nonparallel side is called a leg. Base angles are two consecutive angles whose common side is a base. Copyright by Holt, Rinehart and Winston. 84 Holt Geometry All rights reserved. Name Date Class LESSON 6-6 Review for Mastery Properties of Kites and Trapezoids continued Trapezoid Midsegment Theorem The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. The midsegment of a trapezoid is parallel to each base. AB MN and AB LP The length of the midsegment is one-half the sum of the length of the bases. AB 1 2 (MN LP) Find each value so that the trapezoid is isosceles. 6. Find the value of x. 7. AC 2z 9, BD 4z 3. Find the value of z. 2 3 4 1 (5X 2 32) (7X 2 ) . Find each length. 8. KL 9. PQ . ( 16 26 2. 3 4 1 7.5 5.5 0 10. EF 11. WX . 14 4.3 6 9 7 8. 5 35 22.9 0. . - AB is the midsegment of LMNP. Copyright by Holt, Rinehart and Winston. 85 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Ratio and Proportion 7-1 A ratio is a comparison of two numbers by division. Ratios can be written in various forms. Ratios comparing x and y Ratios comparing 3 and 2 x to y x. y x y. where y 0 3 to 2 3. 2 3 2 Slope is a ratio that compares the rise, or change in y, to the run, or change in x. Slope rise run y 2 y 1 x 2 x 1 Definition of slope 5 3 7 3 Substitution 2 4 or 1 2 Simplify. A ratio can involve more than two numbers. The ratio of the angle measures in a triangle is 2. 3. 4. What is the measure of the smallest angle Let the angle measures be 2x , 3x , and 4x . 2x 3x 4x 180 Triangle Sum Theorem 9x 180 Simplify. x 20 Divide both sides by 9. The smallest angle measures 2x. So 2x 2(20) 40. Write a ratio expressing the slope of each line. 1. X Y. 3 3 0 3 3 2. X Y 3 3 0 3 3 3. X Y 3 3 0 3 3 4. The ratio of the side lengths of a triangle 5. The ratio of the angle measures in a triangle is 2. 4. 5, and the perimeter is 55 cm. is 7. 13. 16. What is the measure of the What is the length of the shortest side largest angle X Y (3, 3) (7, 5) 3 3 0 X X X smallest angle Copyright by Holt, Rinehart and Winston. 86 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Ratio and Proportion continued 7-1 A proportion is an equation stating that two ratios are equal. In every proportion, the product of the extremes equals the product of the means. Cross Products Property In a proportion, if a b c d and b and d 0, then ad bc. A B C D You can solve a proportion like x 8 35 56 by finding the cross products. x 8 35 56 x(56) 8(35) Cross Products Property 56x 280 Simplify. x 5 Divide both sides by 56. You can use properties of proportions to find ratios. Given that 8a 6b, find the ratio of a to b in simplest form. 8a 6b a b 6 8 Divide both sides by b. a b 3 4 Simplify 6 8. The ratio of a to b in simplest form is 3 to 4. Solve each proportion. 6. 9 t 36 28 7. x 32 15 16 8. 24 42 y 7 9. 2a 3 8 3a 10. Given that 5b 20c, find the ratio b c in 11. Given that 24x 9y, find the ratio x. y simplest form. in simplest form. a and d are the extremes. b and c are the means. means a. b c. d extremes Copyright by Holt, Rinehart and Winston. 87 Holt Geometry All rights reserved. Name Date Class LESSON 7-2 Review for Mastery Ratios in Similar Polygons Similar polygons are polygons that have the same shape but not necessarily the same size. Similar Polygons . ABC DEF Corresponding angles are congruent. A D B E C F Corresponding sides are proportional. AB DE 6 3 2 BC EF 9 4.5 2 CA FD 10 5 2 A similarity ratio is the ratio of the lengths of the corresponding sides. So, for the similarity statement ABC DEF, the similarity ratio is 2. For DEF ABC, the similarity ratio is 1 2. Identify the pairs of congruent angles and corresponding sides. 1. . Aufrechtzuerhalten. 0 24 12 18 27 36 16 2. 1 2 . 3 4 12.3 7 8 6 3 4 6.15 3.5 Determine whether the polygons are similar. If so, write the similarity ratio and a similarity statement. 3. EFG and HJK 4. rectangles QRST and UVWX ( 18 15 12 10 17 101 101 25.5 8 5 7 6 24 15 2 3 1 4 8 5 Copyright by Holt, Rinehart and Winston. 88 Holt Geometry All rights reserved. Name Date Class LESSON 7-2 Review for Mastery Ratios in Similar Polygons continued You can use properties of similar polygons to solve problems. Rectangle DEFG rectangle HJKL. What is the length of HJKL . ( 40 in. 27 in. 18 in. length of DEFG length of HJKL width of DEFG width of HJKL Write a proportion. 40 x 27 18 Substitute the known values. 40(18) 27(x) Cross Products Property 720 27x Simplify. 26 2 3 x Divide both sides by 27. The length of HJKL is 26 2 3 in. 5. A rectangle is 3.2 centimeters wide and 6. Rectangle CDEF rectangle GHJK, and 8 centimeters long. A similar rectangle is the similarity ratio of CDEF to GHJK is 5 centimeters long. What is the width of 1 16. If DE 20, what is HK the second rectangle 7. ABC is similar to DEF. 8. The two rectangles are similar. What is What is EF the value of x to the nearest tenth . 12 15 19 30.4 19.2 X 12.5 12 18.6 9. MNP QRS, and the ratio 10. Triangle HJK has side lengths 21, 17, and 25. The two shortest sides of triangle WXY have lengths 48.3 and 39.1. If HJK WXY, what is the length of the third side of WXY of MNP to QRS is 5. 2. If MN 42 meters, what is QR Copyright by Holt, Rinehart and Winston. 89 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Triangle Similarity: AA, SSS, and SAS 7-3 Angle-Angle (AA) Similarity If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. . 78 78 57 57 ABC DEF Side-Side-Side (SSS) Similarity If the three sides of one triangle are proportional to the three corresponding sides of another triangle, then the triangles are similar. . 14.4 18 12 10 12 15 ABC DEF Side-Angle-Side (SAS) Similarity If two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar. . 57 57 18 12 10 15 ABC DEF Explain how you know the triangles are similar, and write a similarity statement. 1. 4 6 5 3 1 2 92 92 39 49 2. . ( 18 16 27 24 3. Verify that ABC MNP. . - 0. 8 12 12 10 15 15 Copyright by Holt, Rinehart and Winston. 90 Holt Geometry All rights reserved. Name Date Class LESSON You can use AA Similarity, SSS Similarity, and SAS Similarity to solve problems. First, prove that the triangles are similar. Then use the properties of similarity to find missing measures. Explain why ADE ABC and then find BC. . 3.5 3 2 2 3 Step 1 Prove that the triangles are similar. A A by the Reflexive Property of . AD AB 3 6 1 2 AE AC 2 4 1 2 Therefore, ADE ABC by SAS . Step 2 Find BC. AD AB DE BC Corresponding sides are proportional. 3 6 3.5 BC Substitute 3 for AD, 6 for AB, and 3.5 for DE. 3(BC) 6(3.5) Cross Products Property 3(BC) 21 Simplify. BC 7 Divide both sides by 3. Explain why the triangles are similar and then find each length. 4. GK 5. US ( 12 11 8 5 3 6 7 4 42 28 26 7-3 Review for Mastery Triangle Similarity: AA, SSS, and SAS continued Copyright by Holt, Rinehart and Winston. 91 Holt Geometry All rights reserved. Name Date Class LESSON 7-4 Review for Mastery Applying Properties of Similar Triangles Triangle Proportionality Theorem Example If a line parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. 8 9 You can use the Triangle Proportionality Theorem to find lengths of segments in triangles. Find EG. EG GF DH HF Triangle Proportionality Theorem EG 6 7.5 5 Substitute the known values. EG(5) 6(7.5) Cross Products Property 5(EG) 45 Simplify. EG 9 Divide both sides by 5. Converse of the Triangle Proportionality Theorem Example If a line divides two sides of a triangle proportionally, then it is parallel to the third side. 8 9 Find the length of each segment in Exercises 1 and 2. 1. RQ. 2 3 0 1 7 6 12 2. JN . - 38 20 16 3. Show that TU and WX are parallel. ( 6 5 7.5 6 7 4 5 8 6 45 15 18 XY AC So BX XA BY YC. BX XA BY YC 3 XY AC Copyright by Holt, Rinehart and Winston. 92 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Applying Properties of Similar Triangles continued 7-4 Triangle Angle Bisector Theorem Example An angle bisector of a triangle divides the opposite side into two segments whose lengths are proportional to the lengths of the other two sides. ( Bisector Thm.) 9 . 24 15 40 9 Find LP and LM. LP PN ML NM Bisector Thm. x 6 x 3 10 Substitute the given values. -. 0. 6 X 10 X 3 x(10) 6(x 3) Cross Products Property 10x 6x 18 Distributive Property 4x 18 Simplify. x 4.5 Divide both sides by 4. LP x 4.5 LM x 3 4.5 3 7.5 Find the length of each segment. 4. EF and FG 5. RV and TV X 8 12 X 2 2 3 4 6 3Y 40 16 Y 3 6. NP and LP 7. JK and LK. 0 - 4 5 X 1 X 3 (. 21 14 Y 1 2Y 4 BY YC 15 9 5 3 AB AC 40 24 5 3 Copyright by Holt, Rinehart and Winston. 93 Holt Geometry All rights reserved. Name Date Class LESSON 7-5 Review for Mastery Using Proportional Relationships A scale drawing is a drawing of an object that is smaller or larger than the objects actual size. The drawings scale is the ratio of any length in the drawing to the actual length of the object. The scale for the diagram of the doghouse is 1 in. 3 ft. Find the length of the actual doghouse. 0.75 in. First convert to equivalent units: 1 in. 36 in. (3 ft 12 in./ft). diagram length 1 0.75 diagram length actual length 36 x actual length 1x 36(0.75) Cross Products Property x 27 in. Simplify. The actual length of the doghouse is 27 in. or 2 ft 3 in. The scale of the cabin shown in the blueprint is 1 cm. 2 m. Find the actual lengths of the following walls. 1. HG 2. GL 3. HJ 4. JM A rectangular fitness room in a recreation center is 45 feet long and 28 feet wide. Find the length and width for a scale drawing of the room, using the following scales. 5. 1 in. 1 ft 6. 1 in. 2 ft 7. 1 in. 3 ft 8. 1 in. 6 ft 8 in. ( -. Copyright by Holt, Rinehart and Winston. 94 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Using Proportional Relationships continued 7-5 Proportional Perimeters and Areas Theorem If two figures are similar and their similarity ratio is a b. then the ratio of their perimeters is a b and the ratio of their areas is a b 2. perimeter of ABC perimeter of DEF 36 12 3 1 area of ABC area of DEF 54 6 9 1 3 1 2 AB DE BC EF CA FD 3 1 HJK LMN. The perimeter of HJK is 30 inches, and the area of HJK is 36 square inches. Find the perimeter and area of LMN. The similarity ratio of HJK to LMN 9 12 3 4. perimeter of HJK perimeter of LMN 3 4 The ratio of the perimeters equals the similarity ratio. 30 P 3 4 Substitute the known values. 30(4) P(3) Cross Products Property 40 P Simplify. area of HJK area of LMN 3 4 2 The ratio of the areas equals the square of the similarity ratio. 36 A 9 16 Substitute the known values. 36(16) A(9) Cross Products Property 64 A Simplify. The perimeter of LMN is 40 in. and the area is 64 in 2. 9. PQRS TUVW. Find the perimeter 10. EFG HJK. Find the perimeter and area of TUVW. and area of HJK. 21 cm P 72 cm A 315 cm 2 Q R S P 14 cm U V W T M M 0M M ( . 9 12 15 4 5 3 NN ( IN . IN -. Copyright by Holt, Rinehart and Winston. 95 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Dilations and Similarity in the Coordinate Plane 7-6 A dilation is a transformation that changes the size of a figure but not its shape. The preimage and image are always similar. A scale factor describes how much a figure is enlarged or reduced. Triangle ABC has vertices A(0, 0), B(2, 6), and C(6, 4). Find the coordinates of the vertices of the image after a dilation with a scale factor 1 2. Preimage Image ABC ABC A(0, 0) 0 1 2. 0 1 2 A(0, 0) B(2, 6) 2 1 2. 6 1 2 B(1, 3) C(6, 4) 6 1 2. 4 1 2 C(3, 2) FEG HEJ. Find the coordinates of F and the scale factor. FE HE EG EJ Write a proportion. FE 6 4 8 HE 6, EG 4, and EJ 8. 8(FE) 24 Cross Products Property FE 3 Divide both sides by 8. So the coordinates of F are (0, 3). Since F(0, 3) (0 2, 3 2) H(0, 6), the scale factor is 2 1. 1. Triangle EFG has vertices E(0, 0), F(1, 5), 2. Rectangle LMNP has vertices L(6, 0), and G(7, 2). Find the coordinates of the M(6, 0), N(6, 3), and P(6, 3). Find image after a dilation with a scale factor 2 1. the coordinates of the image after a dilation with a scale factor 1 3. 3. Given that AEB CED, find the 4. Given that LKM NKP, find the coordinates of C and the scale factor. coordinates of P and the scale factor. X Y (3, 0) (9, 0) (0, 0) (0, 2) X Y 0 -(6, 0) (0, 0) ,(0, 9) .(0, 12) X Y (0, 0) (2, 6) (6, 4) (0, 0) (1, 3) (3, 2) 5 4 0 ABC ABC X Y ((0, 6) (4, 0) (8, 0) (0, 0) Copyright by Holt, Rinehart and Winston. 96 Holt Geometry All rights reserved. Name Date Class LESSON 7-6 Review for Mastery Dilations and Similarity in the Coordinate Plane continued You can prove that triangles in the coordinate plane are similar by using the Distance Formula to find the side lengths. Then apply SSS Similarity or SAS Similarity. Use the figure to prove that ABC ADE. Step 1 Determine a plan for proving the triangles similar. A A by the Reflexive Property. If AB AD AC AE. then the triangles are similar by SAS . Step 2 Use the Distance Formula to find the side lengths. AB (1 3) 2 (4 1) 2 AC (5 3) 2 (3 1) 2 13 8 2 2 AD (1 3) 2 (7 1) 2 AE (7 3) 2 (5 1) 2 52 2 13 32 4 2 Step 3 Compare the corresponding sides to determine whether they are proportional. AB AD 13 2 13 1 2 AC AE 2 2 4 2 1 2 The similarity ratio is 1 2. and AB AD AC AE. So ABC ADE by SAS . 5. Prove that FGH FLM. 6. Prove that QRS TUV. X Y -(5, 1) ,(2, 0) (4, 1) (2, 2) ((7, 5) 4 6 2 2 X Y 6(1, 1) 5(0, 3) 1(4, 0) 4(2, 0) 2(0, 6) 3(2, 2) 4 4 4 X Y (3, 1) (1, 4) (5, 3) (1, 7) (7, 5) 5 4 2 0 Copyright by Holt, Rinehart and Winston. 97 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Similarity in Right Triangles 8-1 Altitudes and Similar Triangles The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle. . Similarity statement: ABC ADB BDC The geometric mean of two positive numbers is the positive square root of their product. Find the geometric mean of 5 and 20. Let x be the geometric mean. x 2 (5)(20) Definition of geometric mean x 2 100 Simplify. x 10 Find the positive square root. So 10 is the geometric mean of 5 and 20. Write a similarity statement comparing the three triangles in each diagram. 1. 0 - 2. ( Find the geometric mean of each pair of numbers. If necessary, give the answer in simplest radical form. 3. 3 and 27 4. 9 and 16 5. 4 and 5 6. 8 and 12 original triangle. a x x b x 2 ab x ab x is the geometric mean of a and b. Copyright by Holt, Rinehart and Winston. 98 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Similarity in Right Triangles continued 8-1 You can use geometric means to find side lengths in right triangles. Geometric Means Words Symbols Examples The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the two segments of the hypotenuse. H Y X h 2 xy 6 9 X h 2 xy 6 2 x(9) 36 9x 4 x The length of a leg of a right triangle is the geometric mean of the length of the hypotenuse and the segment of the hypotenuse adjacent to that leg. A B X Y C a 2 xc b 2 yc A 4 9 13 a 2 xc a 2 4(13) a 2 52 a 52 2 13 Find x, y, and z. 7. Y Z X 2 4 8. Y Z X 6 3 9. Y Z X 9 6 10. Y Z X 8qi 2 4 Copyright by Holt, Rinehart and Winston. 99 Holt Geometry All rights reserved. Name Date Class LESSON 8-2 Review for Mastery Trigonometric Ratios Trigonometric Ratios sin A leg opposite A hypotenuse 4 5 0.8 cos A leg adjacent to A hypotenuse 3 5 0.6 tan A leg opposite A leg adjacent to A 4 3 1.33 You can use special right triangles to write trigonometric ratios as fractions. sin 45 sin Q leg opposite Q hypotenuse 2 3 X X 45 45 1 Xqi 2 5 6 X 2X 30 60 4 Xqi 3 x x 2 1 2 2 2 So sin 45 2 2. Write each trigonometric ratio as a fraction and as a decimal ( 15 17 8 rounded to the nearest hundredth. 1. sin K 2. cos H 3. cos K 4. tan H Use a special right triangle to write each trigonometric ratio as a fraction. 5. cos 45 6. tan 45 7. sin 60 8. tan 30 . hypotenuse leg opposite A leg adjacent to A Copyright by Holt, Rinehart and Winston. 100 Holt Geometry All rights reserved. Name Date Class LESSON 8-2 Review for Mastery Trigonometric Ratios continued You can use a calculator to find the value of trigonometric ratios. cos 38 0.7880107536 or about 0.79 You can use trigonometric ratios to find side lengths of triangles. Find WY. cos W adjacent leg hypotenuse Write a trigonometric ratio that involves WY. cos 39 7.5 cm WY Substitute the given values. 8 7 9 7.5 cm 39 WY 7.5 cos 39 Solve for WY. WY 9.65 cm Simplify the expression. Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. 9. sin 42 10. cos 89 11. tan 55 12. sin 6 Find each length. Round to the nearest hundredth. 13. DE 14. FH 18 m 27 ( 10 in. 31 15. JK 16. US. 34.6 mm 18 4 3 5 22.5 cm 66 Copyright by Holt, Rinehart and Winston. 101 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Solving Right Triangles 8-3 Use the trigonometric ratio sin A 0.8 to determine which angle of the triangle is A. sin 1 leg opposite 1 hypotenuse sin 2 leg opposite 2 hypotenuse 6 10 8 10 0.6 0.8 Since sin A sin 2, 2 is A. If you know the sine, cosine, or tangent of an acute angle measure, then you can use your calculator to find the measure of the angle. Inverse Trigonometric Functions Symbols Examples sin A x sin 1 x mA sin 30 1 2 sin 1 1 2 30 cos B x cos 1 x mB cos 45 2 2 cos 1 2 2 45 tan C x tan 1 x mC tan 76 4.01 tan 1 4.01 76 Use the given trigonometric ratio to determine which angle of the triangle is A. 1. sin A 1 2 2. cos A 13 15 3. cos A 0.5 4. tan A 15 26 Use your calculator to find each angle measure to the nearest degree. 5. sin 1 (0.8) 6. cos 1 (0.19) 7. tan 1 (3.4) 8. sin 1 1 5 Copyright by Holt, Rinehart and Winston. 102 Holt Geometry All rights reserved. Name Date Class LESSON To solve a triangle means to find the measures of all the angles and all the sides of the triangle. Find the unknown measures of JKL. MM . Step 1: Find the missing side lengths. sin 38 JL leg opposite K 22 hypotenuse 13.54 mm JL JL 2 LK 2 JK 2 Pythagorean Theorem 13.542 LK 2 22 2 Substitute the known values. LK 17.34 mm Simplify. Step 2: Find the missing angle measures. mJ 90 38 Acute of a rt. are complementary. 52 Simplify. So JL 13.54 mm, LK 17.34 mm, and mJ 52. Find the unknown measures. Round lengths to the nearest hundredth and angle measures to the nearest degree. 9. 10 ft . 53 10. 8.2 mi 4 mi ( 11. 14 km 56 3 2 1 12. 31 cm 36 cm 8 7 9 For each triangle, find the side lengths to the nearest hundredth and the angle measures to the nearest degree. 13. M(5, 1), N(1, 1), P(5, 7) 14. J(2, 3), K(1, 4), L(1, 3) 8-3 Review for Mastery Solving Right Triangles continued Copyright by Holt, Rinehart and Winston. 103 Holt Geometry All rights reserved. Name Date Class LESSON 8-4 Review for Mastery Angles of Elevation and Depression line of sight Classify each angle as an angle of elevation or an angle of depression. 1. 1 2. 2 1 2 Use the figure for Exercises 3 and 4. Classify each angle as an angle of elevation or an angle of depression. 3. 3 3 4 4. 4 Use the figure for Exercises 58. Classify each angle as an angle of elevation or an angle of depression. 5. 1 6. 2 3 4 1 2 7. 3 8. 4 An angle of depression is formed by a horizontal line and a line of sight below it. An angle of elevation is formed by a horizontal line and a line of sight above it. Copyright by Holt, Rinehart and Winston. 104 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Angles of Elevation and Depression continued 8-4 You can solve problems by using angles of elevation and angles of depression. Sarah is watching a parade from a 20-foot balcony. The angle of depression to the parade is 47. What is the distance between Sarah and the parade Draw a sketch to represent the given information. Let A represent 47 X 20 ft. Sarah and let B represent the parade. Let x represent the distance between Sarah and the parade. mB 47 by the Alternate Interior Angles Theorem. Write a sine ratio using B. sin 47 20 ft leg opposite B x hypotenuse x sin 47 20 ft Multiply both sides by x. x 20 sin 47 ft Divide both sides by sin 47. 27 ft x Simplify the expression. The distance between Sarah and the parade is about 27 feet. 9. When the angle of elevation to the sun 10. A person snorkeling sees a turtle on the is 52, a tree casts a shadow that is ocean floor at an angle of depression of 9 meters long. What is the height of 38. She is 14 feet above the ocean floor. the tree Round to the nearest tenth How far from the turtle is she Round to of a meter. the nearest foot. 52 9 m X 38 14 ft X 11. Jared is standing 12 feet from a 12. Maria is looking out a 17-foot-high rock-climbing wall. When he looks up window and sees two deer. The angle of to see his friend ascend the wall, the depression to the deer is 26. What is the angle of elevation is 56. How high up horizontal distance from Maria to the the wall is his friend Round to the deer Round to the nearest foot. nearest foot. Copyright by Holt, Rinehart and Winston. 105 Holt Geometry All rights reserved. Name Date Class LESSON 8-5 Review for Mastery Law of Sines and Law of Cosines You can use a calculator to find trigonometric ratios for obtuse angles. sin 115 0.906307787 cos 270 0 tan 96 9.514364454 The Law of Sines For any ABC with side lengths a, b, and c that are opposite angles A, B, and C, respectively, sin A a sin B b sin C c. A C B Find mP. Round to the nearest degree. sin P MN sin N PM Law of Sines sin P 10 in. sin 36 7 in. MN 10, mN 36, PM 7 sin P 10 in. sin 36 7 in. Multiply both sides by 10 in. sin P 0.84 Simplify. mP sin 1 (0.84) Use the inverse sine function to find mP. mP 57 Simplify. Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. 1. cos 104 2. tan 225 3. sin 100 Find each measure. Round the length to the nearest tenth and the angle measure to the nearest degree. 4. TU 5. mE 4 3 5 64 41 18 m 102 42 in. 26 in. - 0. 10 in. 36 7 in. Copyright by Holt, Rinehart and Winston. 106 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Law of Sines and Law of Cosines continued 8-5 The Law of Cosines For any ABC with side lengths a, b, and c that are opposite angles A, B, and C, respectively, a 2 b 2 c 2 2bc cos A, b 2 a 2 c 2 2ac cos B, c 2 a 2 b 2 2ab cos C. A C B Find HK. Round to the nearest tenth. HK 2 HJ 2 JK 2 2(HJ)(JK) cos J Law of Cosines 289 196 2(17)(14) cos 50 Substitute the known values. ( 50 17 ft 14 ft HK 2 179.0331 ft 2 Simplify. HK 13.4 ft Find the square root of both sides. You can use the Law of Sines and the Law of Cosines to solve triangles according to the information you have. Use the Law of Sines if you know Use the Law of Cosines if you know two angle measures and any side length, or two side lengths and a nonincluded angle measure two side lengths and the included angle measure, or three side lengths Find each measure. Round lengths to the nearest tenth and angle measures to the nearest degree. 6. EF 7. mX 43 10 cm 9 cm 7 9 8 7 6 8 8. mR 9. AB 3 4 2 21 mi 15 mi 95 . 11 km 16 km 28 Copyright by Holt, Rinehart and Winston. 107 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Vectors 8-6 A vector is a quantity that has both length and direction. The vector below may be named u HJ or u v. ( length of V The component form of a vector lists the horizontal and vertical change from the initial point to the terminal point. x, y So the component form of u AB is 3, 4. You can also find the component form of a vector if you know the coordinates of the vector. Suppose u JK has coordinates J(6, 0) and K(1, 3). u JK x 2 x 1. y 2 y 1 Subtract the coordinates of the initial point from the coordinates of the terminal point. u JK 1 6, 3 0 Substitute the coordinates of points J and K. u JK 5, 3 Simplify. The component form of u JK is 5, 3. Write each vector in component form. 1. u FG 2. u QR X Y 2 2 0 X 1 2 Y 2 2 0 3. u LM with initial point L(6, 2) and terminal 4. The vector with initial point C(0, 5) and point M(1, 5) terminal point D(2, 3) J is the terminal point. H is the initial point. horizontal change from initial point vertical change from initial point left 3 units up 4 units. Copyright by Holt, Rinehart and Winston. 108 Holt Geometry All rights reserved. Name Date Class LESSON 8-6 Review for Mastery Vectors continued The magnitude of a vector is its length. The magnitude of u AB is written u AB . The direction of a vector is the angle that it makes with a horizontal line, such as the x-axis. Draw the vector 5, 2 on a coordinate plane. Find its magnitude and direction. To draw the vector, use the origin as the initial point. Then (5, 2) is the terminal point. Use the Distance Formula to find the magnitude. 5, 2 (5 0) 2 (2 0) 2 29 5.4 To find the direction, draw right triangle ABC. Then find the measure of A. tan A 2 5 mA tan 1 2 5 22 Find the magnitude of each vector to the nearest tenth. 5. 3, 1 6. 4, 6 Draw each vector on a coordinate plane. Find the direction of each vector to the nearest degree. 7. 4, 4 8. 6, 3 X Y X Y Equal vectors have the same magnitude and the same direction. Parallel vectors have the same direction or have opposite directions. Identify each of the following. 9. equal vectors A C B D 10. parallel vectors X Y X. Y Copyright by Holt, Rinehart and Winston. 109 Holt Geometry All rights reserved. Name Date Class LESSON Area of Triangles and Quadrilaterals Parallelogram H B A bh Triangle H B A 1 2 bh Trapezoid H B 2 B 1 A 1 2 (b 1 b 2 )h Find the perimeter of the rectangle in which A 27 mm 2. 3 mm Step 1 Find the height. A bh Area of a rectangle 27 3h Substitute 27 for A and 3 for b. 9 mm h Divide both sides by 3. Step 2 Use the base and the height to find the perimeter. P 2b 2h Perimeter of a rectangle P 2(3) 2(9) 24 mm Substitute 3 for b and 9 for h. Find each measurement. 1. the area of the parallelogram 2. the base of the rectangle in which A 136 mm 2 10 in. 6 in. 8 mm 3. the area of the trapezoid 4. the height of the triangle in which A 192 cm 2 15 m 11 m 7 m 24 cm 5. the perimeter of a rectangle in which 6. b 2 of a trapezoid in which A 5 ft 2. A 154 in 2 and h 11 in. h 2 ft, and b 1 1 ft 9-1 Review for Mastery Developing Formulas for Triangles and Quadrilaterals Copyright by Holt, Rinehart and Winston. 110 Holt Geometry All rights reserved. Name Date Class LESSON Area of Rhombuses and Kites Rhombus D 1 D 2 A 1 2 d 1 d 2 Kite D 1 D 2 A 1 2 d 1 d 2 Find d 2 of the kite in which A 156 in 2. A 1 2 d 1 d 2 Area of a kite 156 1 2 (26)d 2 Substitute 156 in 2 for A and 26 in. for d 1. 26 in. 156 13d 2 Simplify. 12 in. d 2 Divide both sides by 13. Find each measurement. 7. the area of the rhombus 8. d 1 of the kite in which A 414 ft 2. 14 cm 10 cm 23 ft 9. d 2 of the rhombus in which A 90 m 2 10. d 1 of the kite in which A 39 mm 2 15 m 6 mm 11. d 1 of a kite in which A 16x m 2 and 12. the area of a rhombus in which d 2 8 m d 1 4ab in. and d 2 7a in. Review for Mastery Developing Formulas for Triangles and Quadrilaterals continued 9-1 Copyright by Holt, Rinehart and Winston. 111 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Developing Formulas for Circles and Regular Polygons Circumference and Area of Circles A circle with diameter d and radius r has circumference C d or C 2r. A circle with radius r has area A r 2. Find the circumference of circle S in which A 81 cm 2. Step 1 Use the given area to solve for r. 3 R cm A r 2 Area of a circle 81 cm 2 r 2 Substitute 81 for A. 81 cm 2 r 2 Divide both sides by . 9 cm r Take the square root of both sides. Step 2 Use the value of r to find the circumference. C 2r Circumference of a circle C 2(9 cm) 18 cm Substitute 9 cm for r and simplify. Find each measurement. 1. the circumference of circle B 2. the area of circle R in terms of D 6 P cm 2 5 m 3. the area of circle Z in terms of 4. the circumference of circle T in terms of . D 22 ft 4 10 in. 5. the circumference of circle X in 6. the radius of circle Y in which C 18 cm which A 49 in 2 9-2 D R Copyright by Holt, Rinehart and Winston. 112 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Developing Formulas for Circles and Regular Polygons continued 9-2 Area of Regular Polygons The area of a regular polygon with apothem a and perimeter P is A 1 2 aP. Find the area of a regular hexagon with side length 10 cm. Step 1 Draw a figure and find the measure of a central angle. Each central angle measure of a regular n-gon is 360 n. 10 cm 5 cm A 60 30 Step 2 Use the tangent ratio to find the apothem. You could also use the 30-60-90 Thm. in this case. tan 30 leg opposite 30 angle leg adjacent to 30 angle Write a tangent ratio. tan 30 5 cm a Substitute the known values. a 5 cm tan 30 Solve for a. Step 3 Use the formula to find the area. A 1 2 aP A 1 2 5 tan 30 60 a 5 tan 30. P 6 10 or 60 cm A 259.8 cm 2 Simplify. Find the area of each regular polygon. Round to the nearest tenth. 7. 12 cm 8. 4 in. 9. a regular hexagon with an apothem of 3 m 10. a regular decagon with a perimeter of 70 ft A The apothem is the distance from the center to a side. The center is equidistant from the vertices. A central angle has its vertex at the center. This central angle measure is 360 n 60. Copyright by Holt, Rinehart and Winston. 113 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Composite Figures The figure at right is called a composite figure because it is made up of simple shapes. To find its area, first find 18 cm 13 cm 11 cm 7 cm the areas of the simple shapes and then add. Divide the figure into a triangle and a rectangle. CM CM CM CM area of triangle: A 1 2 bh area of rectangle: A bh 1 2 (5)(4) 18(7) 10 cm 2 126 cm 2 The area of the whole figure is 10 126 136 cm 2. Find the shaded area. Round to the nearest tenth if necessary. 1. YD YD YD YD 2. MM MM MM MM 3. 16 ft 16 ft 16 ft 4. M M M M M 9-3 The base of the triangle is 18 13 5 cm. The height of the triangle is 11 7 4 cm. Copyright by Holt, Rinehart and Winston. 114 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Composite Figures continued You can also find the area of composite figures by using subtraction. To find the area of the figure at right, 12 in. 4 in. 4 in. 4 in. 4 in. 9 in. 9 in. subtract the area of the square from the area of the rectangle. area of rectangle: area of square: A bh A s 2 12(9) 4 2 108 in 2 16 in 2 The shaded area is 108 16 92 in 2. You can use composite figures to estimate the area of A B an irregular shape like the one shown at right. The grid has squares with side lengths of 1 cm. area of square a: A 2 2 4 cm 2 area of triangle b: A 1 2 (2)(2) 2 cm 2 The shaded area is about 4 2 or 6 cm 2. Find the shaded area. Round to the nearest tenth if necessary. 5. 18 mm 22 mm 9 mm 6. 9 cm 5 cm Use a composite figure to estimate each shaded area. The grid has squares with side lengths of 1 cm. 7. 8. 9-3 Copyright by Holt, Rinehart and Winston. 115 Holt Geometry All rights reserved. Name Date Class LESSON 9-4 Review for Mastery Perimeter and Area in the Coordinate Plane One way to estimate the area of irregular shapes in the coordinate plane is to count the squares on the grid. You can estimate the number of whole squares and the number of half squares and then add. The polygon with vertices A(3, 1), B(3, 3), C(2, 3), and X. Y 2 2 2 2 0 D(4, 1) is drawn in the coordinate plane. The figure is a trapezoid. Use the Distance Formula to find the length of CD. CD (4 2) 2 (1 3) 2 20 2 5 perimeter of ABCD: P AB BC CD DA 4 5 2 5 7 20.5 units area of ABCD: A 1 2 (b 1 b 2 )(h) 1 2 (5 7)(4) 24 units 2 Estimate the area of each irregular shape. 1. X Y 2 2 2 2 0 2. X Y 2 3 3 2 0 Draw and classify each polygon with the given vertices. Find the perimeter and area of each polygon. 3. F(2, 3), G(2, 3), H(2, 0) 4. Q(4, 0), R(2, 4), S(2, 2), T(0, 2) X Y X Y Copyright by Holt, Rinehart and Winston. 116 Holt Geometry All rights reserved. Name Date Class LESSON 9-4 Review for Mastery Perimeter and Area in the Coordinate Plane continued When a figure in a coordinate plane does not have an area formula, another method can be used to find its area. Find the area of the polygon with vertices N(4, 1), P(1, 3), Q(4, 3), and R(2, 2). Step 1 Draw the polygon and enclose it in a rectangle. x R P Q y 2 2 2 3 0 a b c N Step 2 Find the area of the rectangle and the areas of the parts of the rectangle that are not included in the figure. rectangle: A bh 8 5 40 units 2 a: A 1 2 bh 1 2 (3)(4) 6 units 2 b: A 1 2 bh 1 2 (2)(5) 5 units 2 c: A 1 2 bh 1 2 (6)(1) 3 units 2 Step 3 Subtract to find the area of polygon NPQR. A area of rectangle area of parts not included in figure 40 6 5 3 26 units 2 Find the area of each polygon with the given vertices. 5. X Y 2 2 2 2 0 7(3, 1) :(2, 4) 8(3, 4) 9(3, 1) 6. X Y 3 2 2 2 0 6(3, 3) 3(3, 1) 5(4, 0) 4(2, 3) 7. A(1, 1), B(2, 3), C(2, 4), D(4, 1) 8. H(3, 7), J(7, 2), K(4, 0), L(1, 1) Copyright by Holt, Rinehart and Winston. 117 Holt Geometry All rights reserved. Name Date Class LESSON 9-5 Review for Mastery Effects of Changing Dimensions Proportionally What happens to the area of the parallelogram if the base is tripled original dimensions: triple the base: 4 cm 5 cm A bh A bh 4(5) 12(5) 20 cm 2 60 cm 2 Notice that 60 3(20). If the base is multiplied by 3, the area is also multiplied by 3. Describe the effect of each change on the area of the given figure. 1. The length of the rectangle is doubled. 2. The base of the triangle is multiplied by 4. 18 m 7 m 5 in. 3 in. 3. The height of the parallelogram is 4. The width of the rectangle is multiplied multiplied by 5. by 1 2. 3 yd 2 yd 6 ft 4 ft 5. The height of the trapezoid is multiplied by 3. 6. The radius of the circle is multiplied by 1 2. 4 cm 6 cm 11 cm 8 cm Copyright by Holt, Rinehart and Winston. 118 Holt Geometry All rights reserved. Name Date Class LESSON 9-5 Review for Mastery Effects of Changing Dimensions Proportionally continued What happens if both the base and height of the parallelogram are tripled original dimensions: triple the base and height: 4 cm 5 cm A bh A bh 4(5) 12(15) 20 cm 2 180 cm 2 When just the base is multiplied by 3, the area is also multiplied by 3. When both the base and height are multiplied by 3, the area is multiplied by 3 2. or 9. Effects of Changing Dimensions Proportionally Change in Dimensions Perimeter or Circumference Area Consider a rectangle whose length and width w are each multiplied by a. A(W) A() The perimeter changes by a factor of a. P 2 2w new perimeter: P a(2 2w) The area changes by a factor of a 2. original area: A w new area: A a 2 (w) Describe the effect of each change on the perimeter or circumference and the area of the given figure. 7. The side length of the square is 8. The base and height of the rectangle are multiplied by 6. both multiplied by 1 2. 7 cm 6 ft 4 ft 9. The base and height of a triangle with base 10. A circle has radius 5 mm. The radius is 7 in. and height 3 in. are both doubled. multiplied by 4. Copyright by Holt, Rinehart and Winston. 119 Holt Geometry All rights reserved. Name Date Class LESSON 9-6 Review for Mastery Geometric Probability The theoretical probability of an event occurring is P number of outcomes in the event number of outcomes in the sample space. The geometric probability of an event occurring is found by determining a ratio of geometric measures such as length or area. Geometric probability is used when an experiment has an infinite number of outcomes. Finding Geometric Probability Use Length Use Angle Measures A point is chosen randomly on AD. Find the probability that the point is on BD. 2 4 6. P all points on BD all points on AD BD AD 10 12 5 6 Use the spinner to find the probability of the pointer landing on the 160 space. P all points in 160 region all points in circle 160 360 4 9 A point is chosen randomly on EH. Find the 5 1 2 ( probability of each event. 1. The point is on FH. 2. The point is not on EF. 3. The point is on EF or GH. 4. The point is on EG. Use the spinner to find the probability of each event. 90 30 135 80 75 5. the pointer landing on 135 6. the pointer landing on 75 7. the pointer landing on 90 or 75 8. the pointer landing on 30 160 80 120 Copyright by Holt, Rinehart and Winston. 120 Holt Geometry All rights reserved. Name Date Class LESSON 9-6 Review for Mastery Geometric Probability continued You can also use area to find geometric probability. Find the probability that a point chosen randomly inside the rectangle is in the triangle. area of triangle: A 1 2 bh 10 cm 5 cm 6 cm 3 cm 1 2 (6)(3) 9 cm 2 area of rectangle: A bh 10(5) 50 cm 2 P all points in triangle all points in rectangle area of triangle area of rectangle 9 cm 2 50 cm 2 The probability is P 0.18. Find the probability that a point chosen randomly inside the rectangle is in each shape. Round to the nearest hundredth. 9. the square 10. the triangle 7 in. 4 in. 2 in. 14 cm 13 cm 12 cm 5 cm 11. the circle 12. the regular pentagon MM MM MM 10 ft 8 ft 4 ft Copyright by Holt, Rinehart and Winston. 121 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Solid Geometry 10-1 Three-dimensional figures, or solids, can have flat or curved surfaces. Prisms and pyramids are named by the shapes of their bases. Solids Prisms Pyramids bases bases triangular rectangular triangular rectangular prism prism pyramid pyramid Cylinder Cone bases base vertex Neither cylinders nor cones have edges. Classify each figure. Name the vertices, edges, and bases. 1. 2 4 1 3 2. 3. ( 4. - Each flat surface is called a face. A vertex is the point where three or more faces intersect. In a cone, it is where the curved surface comes to a point. An edge is the segment where two faces intersect. Copyright by Holt, Rinehart and Winston. 122 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Solid Geometry continued 10-1 A net is a diagram of the surfaces of a three-dimensional figure. It can be folded to form the three-dimensional figure. The net at right has one rectangular face. The remaining faces are triangles, and so the net forms a rectangular pyramid. A cross section is the intersection of a three-dimensional figure and a plane. Describe the three-dimensional figure that can be made from the given net. 5. 6. Describe each cross section. 7. 8. rectangular pyramid net of rectangular pyramid The cross section is a triangle. Copyright by Holt, Rinehart and Winston. 123 Holt Geometry All rights reserved. Name Date Class LESSON 10-2 Review for Mastery Representations of Three-Dimensional Figures An orthographic drawing of a three-dimensional object shows six different views of the object. The six views of the figure at right are shown below. Top: Bottom: Front: Back: Left: Right: Draw all six orthographic views of each object. Assume there are no hidden cubes. 1. 2. Copyright by Holt, Rinehart and Winston. 124 Holt Geometry All rights reserved. Name Date Class LESSON 10-2 Review for Mastery Representations of Three-Dimensional Figures continued An isometric drawing is drawn on isometric dot paper and shows three sides of a figure from a corner view. A solid and an isometric drawing of the solid are shown. In a one-point perspective drawing, nonvertical lines are drawn so that they meet at a vanishing point. You can make a one-point perspective drawing of a triangular prism. Draw an isometric view of each object. Assume there are no hidden cubes. 3. 4. Draw each object in one-point perspective. 5. a triangular prism with bases 6. a rectangular prism that are obtuse triangles Step 2 From each vertex of the triangle, draw dashed segments to the vanishing point. Step 4 Draw the edges of the prism. Use dashed lines for hidden edges. Erase segments that are not part of the prism. Step 1 Draw a horizontal line and a vanishing point on the line. Draw a triangle below the line. Step 3 Draw a smaller triangle with vertices on the dashed segments. Copyright by Holt, Rinehart and Winston. 125 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Formulas in Three Dimensions 10-3 A polyhedron is a solid formed by four or more polygons that intersect only at their edges. Prisms and pyramids are polyhedrons. Cylinders and cones are not. Eulers Formula For any polyhedron with V vertices, E edges, and F faces, V E F 2. Example V E F 2 Eulers Formula 4 6 4 2 V 4, E 6, F 4 2 2 4 vertices, 6 edges, 4 faces Diagonal of a Right Rectangular Prism The length of a diagonal d of a right rectangular prism with length , width w, and height h is d 2 w 2 h 2. Find the height of a rectangular prism with a 4 cm by 3 cm base and a 7 cm diagonal. d 2 w 2 h 2 Formula for the diagonal of a right rectangular prism 7 4 2 3 2 h 2 Substitute 7 for d, 4 for , and 3 for w. 49 4 2 3 2 h 2 Square both sides of the equation. 24 h 2 Simplify. 4.9 cm h Take the square root of each side. Find the number of vertices, edges, and faces of each polyhedron. Use your results to verify Eulers Formula. 1. 2. Find the unknown dimension in each figure. Round to the nearest tenth if necessary. 3. the length of the diagonal of a 6 cm 4. the height of a rectangular prism with a by 8 cm by 11 cm rectangular prism 4 in. by 5 in. base and a 9 in. diagonal H W D Copyright by Holt, Rinehart and Winston. 126 Holt Geometry All rights reserved. Name Date Class LESSON A three-dimensional coordinate system has three perpendicular axes: x-axis y-axis z-axis An ordered triple (x, y, z) is used to locate a point. The point at (3, 2, 4) is graphed at right. Formulas in Three Dimensions Distance Formula The distance between the points (x 1. y 1. z 1 ) and (x 2. y 2. z 2 ) is d (x 2 x 1 ) 2 (y 2 y 1 ) 2 (z 2 z 1 ) 2. Midpoint Formula The midpoint of the segment with endpoints (x 1. y 1. z 1 ) and (x 2. y 2. z 2 ) is M x 1 x 2 2. y 1 y 2 2. z 1 z 2 2 . Find the distance between the points (4, 0, 1) and (2, 3, 0). Find the midpoint of the segment with the given endpoints. d (x 2 x 1 ) 2 (y 2 y 1 ) 2 (z 2 z 1 ) 2 Distance Formula (2 4) 2 (3 0) 2 (0 1) 2 (x 1. y 1. z 1 ) (4, 0, 1), (x 2. y 2. z 2 ) (2, 3, 0) 4 9 1 Simplify. 14 3.7 units Simplify. The distance between the points (4, 0, 1) and (2, 3, 0) is about 3.7 units. M x 1 x 2 2. y 1 y 2 2. z 1 z 2 2 M 4 2 2. 0 3 2. 1 0 2 Midpoint Formula M(3, 1.5, 0.5) Simplify. The midpoint of the segment with endpoints (4, 0, 1) and (2, 3, 0) is M(3, 1.5, 0.5). Find the distance between the given points. Find the midpoint of the segment with the given endpoints. Round to the nearest tenth if necessary. 5. (0, 0, 0) and (6, 8, 2) 6. (0, 6, 0) and (4, 8, 0) 7. (9, 1, 4) and (7, 0, 7) 8. (2, 4, 1) and (3, 3, 5) 10-3 Review for Mastery Formulas in Three Dimensions continued Y Z X (3, 2, 4) 3 2 4 Copyright by Holt, Rinehart and Winston. 127 Holt Geometry All rights reserved. Name Date Class LESSON 10-4 Review for Mastery Surface Area of Prisms and Cylinders The lateral area of a prism is the sum of the areas of all the lateral faces. A lateral face is not a base. The surface area is the total area of all faces. Lateral and Surface Area of a Right Prism Lateral Area The lateral area of a right prism with base perimeter P and height h is L Ph. Surface Area The surface area of a right prism with lateral area L and base area B is S L 2B, or S Ph 2B. The lateral area of a right cylinder is the curved surface that connects the two bases. The surface area is the total area of the curved surface and the bases. Lateral and Surface Area of a Right Cylinder Lateral Area The lateral area of a right cylinder with radius r and height h is L 2rh. Surface Area The surface area of a right cylinder with lateral area L and base area B is S L 2B, or S 2rh 2r 2. Find the lateral area and surface area of each right prism. 1. 9 ft 4 ft 3 ft 2. CM CM CM CM Find the lateral area and surface area of each right cylinder. Give your answers in terms of . 3. 6 in. 5 in. 4. 15 cm 8 cm H lateral face H lateral surface Copyright by Holt, Rinehart and Winston. 128 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Surface Area of Prisms and Cylinders continued 10-4 You can find the surface area of a composite three-dimensional figure like the one shown at right. surface area of small prism surface area of large prism hidden surfaces The dimensions are multiplied by 3. Describe the effect on the surface area. original surface area: new surface area, dimensions multiplied by 3: S Ph 2B S Ph 2B 20(3) 2(16) P 20, h 3, B 16 60(9) 2(144) P 60, h 9, B 144 92 mm 2 Simplify. 828 mm 2 Simplify. Notice that 92 9 828. If the dimensions are multiplied by 3, the surface area is multiplied by 3 2. or 9. Find the surface area of each composite figure. Be sure to subtract the hidden surfaces of each part of the composite solid. Round to the nearest tenth. 5. 2 cm 2 cm 2 cm 2 cm 5 cm 3 cm 6. 2 in. 2 in. 1 in. 4 in. 3 in. Describe the effect of each change on the surface area of the given figure. 7. The length, width, and height are 8. The height and radius are multiplied by 1 2. multiplied by 2. 5 cm 1 cm 2 cm 2 m 4 m 2 cm 2 cm 2 cm 2 cm 5 cm 3 cm 8 mm 2 mm 3 mm Copyright by Holt, Rinehart and Winston. 129 Holt Geometry All rights reserved. Name Date Class LESSON 10-5 Review for Mastery Surface Area of Pyramids and Cones Lateral and Surface Area of a Regular Pyramid Lateral Area The lateral area of a regular pyramid with perimeter P and slant height is L 1 2 P. Surface Area The surface area of a regular pyramid with lateral area L and base area B is S L B, or S 1 2 P B. Lateral and Surface Area of a Right Cone Lateral Area The lateral area of a right cone with radius r and slant height is L r. Surface Area The surface area of a right cone with lateral area L and base area B is S L B, or S r r 2. Find the lateral area and surface area of each regular pyramid. Round to the nearest tenth. 1. 5 ft 5 ft 9 ft 2. 6 m 2 m 3 m qi Find the lateral area and surface area of each right cone. Give your answers in terms of . 3. 3 in. 8 in. 4. 6 cm 15 cm slant height base R slant height base Copyright by Holt, Rinehart and Winston. 130 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Surface Area of Pyramids and Cones continued 10-5 The radius and slant height of the cone at right are doubled. Describe the effect on the surface area. original surface area: new surface area, dimensions doubled: S r r 2 S r r 2 (3)(7) (3) 2 r 3, 7 (6)(14) (6) 2 r 6, 14 30 cm 2 Simplify. 120 cm 2 Simplify. If the dimensions are doubled, then the surface area is multiplied by 2 2. or 4. Describe the effect of each change on the surface area of the given figure. 5. The dimensions are tripled. 6. The dimensions are multiplied by 1 2. 3 ft 2 ft 2 ft 2 m 8 m Find the surface area of each composite figure. 7. Hint: Do not include the base area of 8. Hint: Add the lateral areas of the cones. the pyramid or the upper surface area of the rectangular prism. 6 in. 3 in. 4 in. 7 in. 1 cm 5 cm 3 cm 7 cm 3 cm Copyright by Holt, Rinehart and Winston. 131 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Volume of Prisms and Cylinders 10-6 Volume of Prisms Prism The volume of a prism with base area B and height h is V Bh. Right Rectangular Prism The volume of a right rectangular prism with length , width w, and height h is V wh. Cube The volume of a cube with edge length s is V s 3. Volume of a Cylinder The volume of a cylinder with base area B, radius r, and height h is V Bh, or V r 2 h. Find the volume of each prism. 1. 16 cm 4 cm 9 cm 2. 8 in. 5 in. 3 in. Find the volume of each cylinder. Give your answers both in terms of and rounded to the nearest tenth. 3. 8 mm 10 mm 4. 3 ft 5 ft H H W S H R H R Copyright by Holt, Rinehart and Winston. 132 Holt Geometry All rights reserved. Name Date Class LESSON 10-6 Review for Mastery Volume of Prisms and Cylinders continued The dimensions of the prism are multiplied by 1 3. Describe the effect on the volume. original volume: new volume, dimensions multiplied by 1 3. V wh V wh (12)(3)(6) 12, w 3, h 6 (4)(1)(2) 4, w 1, h 2 216 cm 3 Simplify. 8 cm 3 Simplify. Notice that 216 1 27 8. If the dimensions are multiplied by 1 3. the volume is multiplied by 1 3 3. or 1 27. Describe the effect of each change on the volume of the given figure. 5. The dimensions are multiplied by 2. 6. The dimensions are multiplied by 1 4. 7 in. 5 in. 2 in. MM MM Find the volume of each composite figure. Round to the nearest tenth. 7. 10 m 2 m 3 m 4 m 5 m 8. 2 ft 3 ft 3 ft 2 ft 12 cm 6 cm 3 cm Copyright by Holt, Rinehart and Winston. 133 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Volume of Pyramids and Cones 10-7 Volume of a Pyramid The volume of a pyramid with base area B and height h is V 1 3 Bh. Volume of a Cone The volume of a cone with base area B, radius r, and height h is V 1 3 Bh, or V 1 3 r 2 h. Find the volume of each pyramid. Round to the nearest tenth if necessary. 1. 7 in. 5 in. 3 in. 2. 10 mm 8 mm 8 mm Find the volume of each cone. Give your answers both in terms of and rounded to the nearest tenth. 3. 12 ft 4 ft 4. 11 cm 3 cm H R H Copyright by Holt, Rinehart and Winston. 134 Holt Geometry All rights reserved. Name Date Class LESSON 10-7 Review for Mastery Volume of Pyramids and Cones continued The radius and height of the cone are multiplied by 1 2. Describe the effect on the volume. original volume: new volume, dimensions multiplied by 1 2. V 1 3 r 2 h V 1 3 r 2 h 1 3 (4) 2 (6) r 4, h 6 1 3 (2) 2 (3) r 2, h 3 32 in 3 Simplify. 4 in 3 Simplify. If the dimensions are multiplied by 1 2. then the volume is multiplied by 1 2 3. or 1 8. Describe the effect of each change on the volume of the given figure. 5. The dimensions are doubled. 6. The radius and height are multiplied by 1 3. 5 m 3 m 2 m 18 ft 6 ft Find the volume of each composite figure. Round to the nearest tenth if necessary. 7. 3 cm 5 cm 6 cm 6 cm 8. 4 in. 10 in. 8 in. 4 in. 6 in. Copyright by Holt, Rinehart and Winston. 135 Holt Geometry All rights reserved. Name Date Class LESSON 10-8 Review for Mastery Spheres Volume and Surface Area of a Sphere Volume The volume of a sphere with radius r is V 4 3 r 3. Surface Area The surface area of a sphere with radius r is S 4r 2. Find each measurement. Give your answer in terms of . 1. the volume of the sphere 2. the volume of the sphere 5 mm 16 cm 3. the volume of the hemisphere 4. the radius of a sphere with volume 7776 in 3 2 ft 5. the surface area of the sphere 6. the surface area of the sphere 7 in. 20 m R Copyright by Holt, Rinehart and Winston. 136 Holt Geometry All rights reserved. Name Date Class LESSON 10-8 Review for Mastery Spheres continued The radius of the sphere is multiplied by 1 4. Describe the effect on the surface area. original surface area: new surface area, radius multiplied by 1 4. S 4r 2 S 4r 2 4(16) 2 r 16 4(4) 2 r 4 1024 m 2 Simplify. 64 m 2 Simplify. Notice that 1024 1 16 64. If the dimensions are multiplied by 1 4. the surface area is multiplied by 1 4 2. or 1 16. Describe the effect of each change on the given measurement of the figure. 7. surface area 8. volume The radius is multiplied by 4. The dimensions are multiplied by 1 2. 2 ft 14 cm Find the surface area and volume of each composite figure. Round to the nearest tenth. 9. Hint: To find the surface area, add the 10. Hint: To find the volume, subtract the lateral area of the cylinder, the area of volume of the hemisphere from one base, and the surface area of the the volume of the cylinder. hemisphere. 9 cm 12 cm 7 in. 3 in. 16 m Copyright by Holt, Rinehart and Winston. 137 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Lines That Intersect Circles 11-1 Lines and Segments That Intersect Circles A chord is a segment whose endpoints lie on a circle. A secant is a line that intersects a circle at two points. A tangent is a line in the same plane as a circle that intersects the circle at exactly one point, called the point of tangency. Radii and diameters also intersect circles. Tangent Circles Two coplanar circles that intersect at exactly one point are called tangent circles. points of tangency Identify each line or segment that intersects each circle. 1. ( M 2. . -. Find the length of each radius. Identify the point of tangency and write the equation of the tangent line at that point. 3. X. 0 Y 2 2 2 0 4. X 3 4 Y 3 3 2 0 is a tangent. AB and CD are chords. E is a point of tangency. CD is a secant. Copyright by Holt, Rinehart and Winston. 138 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Lines That Intersect Circles continued 11-1 Theorem Hypothesis Conclusion If two segments are tangent to a circle from the same external point, then the segments are congruent. EF and EG are tangent to C. EF EG In the figure above, EF 2y and EG y 8. Find EF. EF EG 2 segs. tangent to from same ext. pt. segs. . 2y y 8 Substitute 2y for EF and y 8 for EG. y 8 Subtract y from each side. EF 2(8) EF 2y substitute 8 for y. 16 Simplify. The segments in each figure are tangent to the circle. Find each length. 5. BC 6. LM . X X . Y Y - 7. RS 8. JK 3 4 0 Y Y 2 ( X X Copyright by Holt, Rinehart and Winston. 139 Holt Geometry All rights reserved. Name Date Class LESSON 11-2 Review for Mastery Arcs and Chords Arcs and Their Measure A central angle is an angle whose vertex is the center of a circle. An arc is an unbroken part of a circle consisting of two points on a circle and all the points on the circle between them. If the endpoints of an arc lie on a diameter, the arc is a semicircle and its measure is 180. Arc Addition Postulate The measure of an arc formed by two adjacent arcs. is the sum of the measures of the two arcs. m ABC m AB m BC Find each measure. ( . 1. m HJ 3. m CDE 2. m FGH 4. m BCD 5. m LMN 2 1 0 . -. 6. m LNP ADC is a major arc. m ADC 360 mABC 360 93 267 ABC is a central angle. AC is a minor arc m AC mABC 93. Copyright by Holt, Rinehart and Winston. 140 Holt Geometry All rights reserved. Name Date Class LESSON 11-2 Review for Mastery Arcs and Chords continued Congruent arcs are arcs that have the same measure. Congruent Arcs, Chords, and Central Angles . If mBEA mCED, then BA CD. . If BA CD. then BA CD. . If BA CD. then mBEA mCED. Congruent central angles have congruent chords. Congruent chords have congruent arcs. Congruent arcs have congruent central angles. In a circle, if a radius or diameter is perpendicular. to a chord, then it bisects the chord and its arc. Find each measure. 7. QR ST. Find m QR. 8. HLG KLJ. Find GH. 3 4 X X 2 1 . Y Y ( Find each length to the nearest tenth. 9. NP 10. EF. - 0 6 4. ( 9 8 Since AB CD. AB bisects CD and CD. Copyright by Holt, Rinehart and Winston. 141 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Sector Area and Arc Length 11-3 Sector of a Circle A sector of a circle is a region bounded by two. R M radii of the circle and their intercepted arc. The area of a sector of a circle is given by the formula A r 2 m 360 . Segment of a Circle A segment of a circle is a region bounded by an arc and. its chord. area of segment ABC area of sector ABC area of ABC Find the area of each sector. Give your answer in terms of and rounded to the nearest hundredth. 1. sector CDE 2. sector QRS CM 3 2 1 120 9 in. Find the area of each segment to the nearest hundredth. 3. ( IN 4. -. M sector ABC segment ABC Copyright by Holt, Rinehart and Winston. 142 Holt Geometry All rights reserved. Name Date Class LESSON Arc Length Arc length is the distance along an arc measured in linear units. M R. The arc length of a circle is given by the formula L 2r m 360 . Find the arc length of JK. L 2 r m 360 Formula for arc length 2 (9 cm) 84 360 Substitute 9 cm for r and 84 for m. CM 21 5 cm Simplify. 13.19 cm Round to the nearest hundredth. Find each arc length. Give your answer in terms of and rounded to the nearest hundredth. 5. AB 6. WX. IN 8 7 CM 7. QR 8. ST 1 2 20 in. 36 3 4 MM 11-3 Review for Mastery Sector Area and Arc Length continued Copyright by Holt, Rinehart and Winston. 143 Holt Geometry All rights reserved. Name Date Class LESSON 11-4 Review for Mastery Inscribed Angles Inscribed Angle Theorem The measure of an inscribed angle is half the measure of its intercepted arc. mABC 1 2 m AC Inscribed Angles If inscribed angles of . a circle intercept the same arc, then the angles are congruent. ABC and ADC intercept AC. so ABC ADC. An inscribed angle . subtends a semicircle if and only if the angle is a right angle. Find each measure. 1. mLMP and m MN 2. mGFJ and m FH -. 0 36 48 ( 36 110 Find each value. 3. x 4. mFJH 2 1 3 (5X 8) ( (4Z 9) 5Z AC is an intercepted arc. ABC is an inscribed angle. Copyright by Holt, Rinehart and Winston. 144 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Inscribed Angles continued 11-4 Inscribed Angle Theorem If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. . A and C are supplementary. B and D are supplementary. ABCD is inscribed in E. Find mG. Step 1 Find the value of z. mE mG 180 EFGH is inscribed in a circle. 4z 3z 5 180 Substitute the given values. ( (3Z 5) 4Z 7z 175 Simplify. z 25 Divide both sides by 7. Step 2 Find the measure of G. mG 3z 5 3(25) 5 80 Substitute 25 for z. Find the angle measures of each quadrilateral. 5. RSTV 6. ABCD 6 2 4 3 (8Y 8) 7Y 11Y . (4X 12) (6X 3) 10X 7. JKLM 8. MNPQ - . (2Z 2) (Z 25) (Z 17) - 0. 1 (4X 5) (4X 10) (3X 7) Copyright by Holt, Rinehart and Winston. 145 Holt Geometry All rights reserved. Name Date Class LESSON 11-5 Review for Mastery Angle Relationships in Circles If a tangent and a secant (or chord) intersect on a circle at the point of tangency, then the measure of the angle formed is half the measure of its intercepted arc. mABC 1 2 m AB If two secants or chords intersect in the interior of a circle, then the measure of the angle formed is half the sum of the measures of its intercepted arcs. m1 1 2 (m AD m BC ) Find each measure. 1. mFGH 2. m LM 216 ( 64. - 3. mJML 4. mSTR 52 70. . - 99 107 3 1 5 2 4 Chords AB and CD intersect at E. Tangent BC and secant BA intersect at B. Copyright by Holt, Rinehart and Winston. 146 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Angle Relationships in Circles continued 11-5 If two segments intersect in the exterior of a circle, then the measure of the angle formed is half the difference of the measures of its intercepted arcs. A Tangent and a Secant Two Tangents Two Secants. m1 1 2 (m AD m BD ) ( m2 1 2 (m EHG m EG ). - m3 1 2 (m JN m KM ) Find the value of x. Since m PVR m PR 360, m PVR 142 360, X 142 1 2 6 0 and m PVR 218. x 1 2 (m PVR m PR ) 1 2 (218 142) x 38 x 38 Find the value of x. 5. X 2 3 5 4 6. X ( 7. X. -. 0 1 8. X . Copyright by Holt, Rinehart and Winston. 147 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Segment Relationships in Circles 11-6 Chord-Chord Product Theorem If two chords intersect in the interior of a circle, then the products of the lengths of the segments of the chords are equal. AE EB CE ED Find the value of x and the length of each chord. HL LJ KL LM Chord-Chord Product Thm. X ( -. 4 9 6 x HL 4, LJ 9, KL 6, LM x 36 6x Simplify. 6 x Divide each side by 6. HJ 4 9 13 KM 6 x 6 6 12 Find the value of the variable and the length of each chord. 1. Y 4 5 6 3 2 2. X ( 3. Z . -. 4. X. Copyright by Holt, Rinehart and Winston. 148 Holt Geometry All rights reserved. Name Date Class LESSON 11-6 Review for Mastery Segment Relationships in Circles continued A secant segment is a segment of a. secant with at least one endpoint on the circle. An external secant segment is the part of the secant segment that lies in the exterior of the circle. A tangent segment is a segment of a tangent with one endpoint on the circle. If two segments intersect outside a circle, the following theorems are true. Secant-Secant Product Theorem The product of the lengths of one secant segment and its external segment equals the product of the lengths of the other secant segment and its external segment. whole outside whole outside AE BE CE DE. Secant-Tangent Product Theorem The product of the lengths of the secant segment and its external segment equals the length of the tangent segment squared. whole outside tangent 2 AE BE DE 2. Find the value of the variable and the length of each secant segment. 5. 1. 2 3 X 8 4 6 0 6. 4 5 6 8 7 Z 8 9 9 Find the value of the variable. 7. ( X 12 4 8. - Y 9 6 BE is an external secant segment. AE is a secant segment. ED is a tangent segment. Copyright by Holt, Rinehart and Winston. 149 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Circles in the Coordinate Plane 11-7 Equation of a Circle The equation of a circle with center ( h, k) and X Y R (H, K) 0 radius r is (x h) 2 ( y k) 2 r 2. Write the equation of C with center C(2, 1) and radius 6. X Y 6 4 6 4 0 (x h) 2 (y k) 2 r 2 Equation of a circle (x 2) 2 ( y (1)) 2 6 2 Substitute 2 for h, 1 for k, and 6 for r. (x 2) 2 ( y 1) 2 36 Simplify. You can also write the equation of a circle if you know the center and one point on the circle. Write the equation of L that has center L(3, 7) and passes through (1, 7). Step 1 Find the radius. Step 2 Use the equation of a circle. r ( x 2 x 1 ) 2 ( y 2 y 1 ) 2 Distance Formula (x h) 2 ( y k) 2 r 2 Equation of a circle r (1 3) 2 (7 7) 2 Substitution (x 3) 2 ( y 7) 2 2 2 (h, k) (3, 7) r 4 2 Simplify. (x 3) 2 ( y 7) 2 4 Simplify. Write the equation of each circle. 1. X Y 3 3 3 3 0 2. X Y 3 4 4 0 3. T with center T(4, 5) and radius 8 4. B that passes through (3, 6) and has center B(2, 6) Copyright by Holt, Rinehart and Winston. 150 Holt Geometry All rights reserved. Name Date Class LESSON 11-7 Review for Mastery Circles in the Coordinate Plane continued You can use an equation to graph a circle by making a table or by identifying its center and radius. Graph (x 1) 2 (y 4) 2 9. The equation of the given circle can be rewritten. (x h) 2 (y k) 2 r 2 (x 1) 2 ( y (4)) 2 3 2 h 1, k 4, and r 3 The center is at (h, k) or (1, 4), and the radius is 3. X Y 3 2 2 4(1, 4) 0 Plot the point (1, 4). Then graph a circle having this center and radius 3. Graph each equation. 5. (x 1) 2 (y 2) 2 9 6. (x 3) 2 (y 1) 2 4 0 X Y 2 2 2 2 0 X Y 2 2 2 2 7. (x 2) 2 (y 2) 2 9 8. (x 1) 2 (y 3) 2 16 0 X Y 2 2 2 2 0 X Y 2 2 2 Copyright by Holt, Rinehart and Winston. 151 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Reflections 12-1 An isometry is a transformation that does not change the shape or size of a figure. Reflections, translations, and rotations are all isometries. A reflection is a transformation that flips a figure across a line. Reflection Not a Reflection The line of reflection is the perpendicular bisector of each segment joining each point and its image. . Tell whether each transformation appears to be a reflection. 1. 2. Copy each figure and the line of reflection. Draw the reflection of the figure across the line. 3. 4. Copyright by Holt, Rinehart and Winston. 152 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Reflections continued 12-1 Reflections in the Coordinate Plane Across the x-axis Across the y - axis Across the line y x 0 X 2(X, Y) 2(X, Y) Y (x, y) (x, y) 0 X 2(X, Y) 2(X, Y) Y (x, y) (x, y) 0 X 2(Y, X) Y X 2(X, Y) Y (x, y) ( y, x) Reflect FGH with vertices F(1, 4), G(2, 4), and H(4, 1) X Y 2 2 2 2 0 ( ( across the x-axis. The reflection of (x, y) is (x, y). F(1, 4) F(1, 4) G(2, 4) G(2, 4) H(4, 1) H(4, 1) Graph the preimage and image. Reflect the figure with the given vertices across the line. 5. M(2, 4), N(4, 2), P(3, 2) y - axis 6. T(4, 1), U(3, 4), V(2, 3), W(0, 1) x - axis X Y X Y 7. Q(3, 1), R(2, 4), S(2, 1) x-axis 8. A(2, 4), B(1, 1), C(5, 1) y x X Y X Y Copyright by Holt, Rinehart and Winston. 153 Holt Geometry All rights reserved. Name Date Class LESSON 12-2 Review for Mastery Translations A translation is a transformation in which all the points of a figure are moved the same distance in the same direction. Translation Not a Translation A translation is a transformation along a vector such that each segment joining a point and its image has the same length as the vector and is parallel to the vector. AA. BB. and CC have the same length V. . as u v and are parallel to u v. Tell whether each transformation appears to be a translation. 1. 2. Copy each figure and the translation vector. Draw the translation of the figure along the given vector. 3. W 4. U Copyright by Holt, Rinehart and Winston. 154 Holt Geometry All rights reserved. Name Date Class LESSON 12-2 Review for Mastery Translations continued Translations in the Coordinate Plane Horizontal Translation Along Vector a, 0 Horizontal Translation Along Vector 0, b Horizontal Translation Along Vector a, b 0 X 4(X A, Y) 4(X, Y) Y (x, y) (x a, y) 0 X 4(X, Y B) 4(X, Y) Y (x, y) (x, y b) 0 X 4(X A, Y B) 4(X, Y) Y (x, y) (x a, y b) Translate JKL with vertices J(0, 1), K(4, 2), and X Y 1 2 2 1 0 . . L(3, 1) along the vector 4, 2. The image of (x, y) is (x 4, y 2). J(0, 1) J(0 4, 1 2) J(4, 3) K(4, 2) K(4 4, 2 2) K(0, 4) L(3, 1) L(3 4, 1 2) L(1, 1) Graph the preimage and image. Translate the figure with the given vertices along the given vector. 5. E(2, 4), F(3, 0), G(3, 4) 0, 3 6. P(4, 1), Q(1, 3), R(0, 4) 4, 1 X Y X Y 7. A(1, 2), B(1, 0), C(3, 1), D(4, 3) 5, 3 8. G(3, 4), H(4, 3), J(1, 2) 1, 6 X Y X Y Copyright by Holt, Rinehart and Winston. 155 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Rotations 12-3 A rotation is a transformation that turns a figure around a fixed point, called the center of rotation. Rotation Not a Rotation A rotation is a transformation about a point P such that 0 1 2 3 1 2 3 each point and its image are the same distance from P. PQ PQ PR PR PS PS Tell whether each transformation appears to be a rotation. 1. 2. Copy each figure and the angle of rotation. Draw the rotation of the figure about point P by mA. 3. 0. 4. 0. angle of rotation center of rotation Copyright by Holt, Rinehart and Winston. 156 Holt Geometry All rights reserved. Name Date Class LESSON Rotations in the Coordinate Plane By 90 About the Origin By 180 About the Origin 0 X .(Y, X) .(X, Y) 90 Y (x, y) (y, x) 0 X .(X, Y) .(X, Y) 180 Y (x, y) (x, y) Rotate MNP with vertices M(1, 1), N(2, 4), X Y 3 3 3 3 0 -. 0 -. 0 and P(4, 3) by 180 about the origin. The image of (x, y) is (x, y). M(1, 1) M(1, 1) N(2, 4) N(2, 4) P(4, 3) P(4, 3) Graph the preimage and image. Rotate the figure with the given vertices about the origin using the given angle. 5. R(0, 0), S(3, 1), T(2, 4) 90 6. A(0, 0), B(4, 2), C(1, 4) 180 X Y X Y 7. E(0, 3), F(3, 5), G(4, 0) 180 8. U(1, 1), V(4, 2), W(3, 4) 90 X Y X Y 12-3 Review for Mastery Rotations continued Copyright by Holt, Rinehart and Winston. 157 Holt Geometry All rights reserved. Name Date Class LESSON 12-4 Review for Mastery Compositions of Transformations A composition of transformations is one transformation followed by another. A glide reflection is the composition of a translation and a reflection across a line parallel to the vector of the translation. Reflect ABC across line along u v and then translate it parallel to u v. V. . Draw the result of each composition of transformations. 1. Translate HJK along u v and then reflect 2. Reflect DEF across line k and it across line m. then translate it along u u. V ( M U K 3. ABC has vertices A(0, 1), B(3, 4), and 4. QRS has vertices Q(2, 1), R(4, 2), C(3, 1). Rotate ABC 180 about the origin and S(1, 3). Reflect QRS across the and then reflect it across the x-axis. y-axis and then translate it along the vector 1, 3. X Y X Y Reflect ABC across line . Translate the image along u v. Copyright by Holt, Rinehart and Winston. 158 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Compositions of Transformations continued 12-4 Any translation or rotation is equivalent to a composition of two reflections. Composition of Two Reflections To draw two parallel lines of reflection that produce a translation: Draw PP. a segment connecting a preimage point P and its corresponding image point P. Draw the midpoint M of PP. Draw the perpendicular bisectors of PM and PM. To draw two intersecting lines that produce a rotation with center C: Draw PCP, where P is a preimage point and P is its corresponding image point. Draw CX. the angle bisector of PCP. Draw the angle bisectors of PCX and PCX. Copy ABC and draw two lines of reflection that. . produce the translation ABC ABC. Step 1 Draw CC and the midpoint M of CC. . - Step 2 Draw the perpendicular bisectors of CM and CM. . - Copy each figure and draw two lines of reflection that produce an equivalent transformation. 5. translation: 6. rotation with center C: JKL JKL PQR PQR . . 0 1 2 0 1 2 Copyright by Holt, Rinehart and Winston. 159 Holt Geometry All rights reserved. Name Date Class LESSON 12-5 Review for Mastery Symmetry A figure has symmetry if there is a transformation of the figure such that the image and preimage are identical. There are two kinds of symmetry. Line Symmetry The figure has a line of symmetry that divides the figure into two congruent halves. one line of symmetry two lines of symmetry no line symmetry Rotational Symmetry When a figure is rotated between 0 and 360, the resulting figure coincides with the original. The smallest angle through which the figure is rotated to coincide with itself is called the angle of rotational symmetry. The number of times that you can get an identical figure when repeating the degree of rotation is called the order of the rotational symmetry. angle: 180 120 no rotational order: 2 3 symmetry Tell whether each figure has line symmetry. If so, draw all lines of symmetry. 1. 2. Tell whether each figure has rotational symmetry. If so, give the angle of rotational symmetry and the order of the symmetry. 3. 4. Copyright by Holt, Rinehart and Winston. 160 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Symmetry continued 12-5 Three-dimensional figures can also have symmetry. Symmetry in Three Dimensions Description Example Plane Symmetry A plane can divide a figure into two congruent halves. Symmetry About an Axis There is a line about which a figure can be rotated so that the image and preimage are identical. A cone has both plane symmetry and symmetry about an axis. Tell whether each figure has plane symmetry, symmetry about an axis, both, or neither. 5. square pyramid 6. prism 7. triangular pyramid 8. cylinder Copyright by Holt, Rinehart and Winston. 161 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Tessellations 12-6 A pattern has translation symmetry if it can be translated along a vector so that the image coincides with the preimage. A pattern with glide reflection symmetry coincides with its image after a glide reflection. Translation Symmetry Translation Symmetry and Glide Reflection Symmetry A tessellation is a repeating pattern that completely covers a plane with no gaps or overlaps. Tessellation Not a Tessellation Identify the symmetry in each pattern. 1. 2. Copy the given figure and use it to create a tessellation. 3. 4. Copyright by Holt, Rinehart and Winston. 162 Holt Geometry All rights reserved. Name Date Class LESSON 12-6 Review for Mastery Tessellations continued A regular tessellation is formed by congruent regular polygons. A semiregular tessellation is formed by two or more different regular polygons. Regular Tessellation Semiregular Tessellation In a tessellation, the measures of the angles that meet at each vertex must have a sum of 360. 90 90 90 90 360 120 120 120 360 3(60) 2(90) 360 Classify each tessellation as regular, semiregular, or neither. 5. 6. Determine whether the given regular polygon(s) can be used to form a tessellation. If so, draw the tessellation. 7. 8. Copyright by Holt, Rinehart and Winston. 163 Holt Geometry All rights reserved. Name Date Class LESSON Review for Mastery Dilations 12-7 A dilation is a transformation that changes the size of a figure but not the shape. Dilation Not a Dilation A dilation is a transformation in which the lines connecting every point A with its image A all intersect at point P, called the center of dilation. 0. . Tell whether each transformation appears to be a dilation. 1. 2. Copy teach triangle and center of dilation. Draw the image of the triangle under a dilation with the given scale factor. 3. scale factor: 2 4. scale factor: 1 2 0 0 Copyright by Holt, Rinehart and Winston. 164 Holt Geometry All rights reserved. Name Date Class LESSON 12-7 Review for Mastery Dilations continued Dilations in the Coordinate Plane For k 1 For 0 k 1 0 X. . Y (x, y) (kx, ky) 0 X. . Y (x, y) (kx, ky) If k has a negative value, the preimage is rotated by 180. Draw the image of EFG with vertices E(0, 0), F(0, 1), and G(2, 1) under a dilation with a scale factor of 3 and centered at the origin. The image of (x, y) is (3x, 3y). X Y 2 2 2 2 0 E(0, 0) E(0(3), 0(3)) E(0, 0) F(0, 1) F(0(3), 1(3)) F(0, 3) G(2, 1) G(2(3), 1(3)) G(6, 3) Graph the preimage and image. Draw the image of the figure with the given vertices under a dilation with the given scale factor and centered at the origin. 5. 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If you experience shortness of breath, sweaty palms or other signs of stress when you are asked to do a step-by-step geometry proof, relax. Here is a short walk-through of a geometry proof that will help you survive beginning geometry. Read the problem carefully. For the purposes of this step-by-step geometry proof, use the following example: Given that triangle ABC is an equilateral triangle and that line AD bisects line BC, prove that the resulting triangle ABD is a right triangle. Draw an illustration of the problem. Having a picture in front of you when doing a geometry proof really helps organize your thoughts. Consider what you know about each piece of given information. For example, because ABC is an equilateral triangle, all three sides must be the same length. Furthermore, all three angles must be equal as well. Since a triangle contains 180 degrees, then each angle in an equilateral triangle must measure 60 degrees. Moving on to the other piece of given information, since line AD bisects side BC, that makes line segments CD and DB equal in length. Use the facts established by the given information to generate more facts that are useful to your geometric proof. Since the line segments CD and DB are equal in length, that means the angle CAD must be equal to the angle DAB. Extrapolate from the facts to get closer to the solution. Since angle A is 60 degrees, the smaller angles must be one half of 60, or 30 degrees. Given that angle B is 60 degrees and that angle DAB is 30 degrees, this accounts for 90 degrees of a triangle. The remaining 90 degrees must be contained in the angle BDA. Since a right triangle must contain a 90-degree angle, you have just proven that triangle ABD is a right triangle. Write out the step-by-step geometric proof of the problem in a two-column format. In the left hand column, write a statement and in the right hand column, write the proof of the statement. Repeat this process until you have documented all of the steps in your thinking process that resulted in your solution.