Geometry Blizzard Bag Day 3

Transcription

1 Class: Date: Geometry Blizzard Bag Day 3 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Three towns, Maybury, Junesville, and Cyanna, will create one sports center. Where should the center be placed so that it is the same distance from all three towns? a. Treat the towns as vertices of a triangle. The center must be placed at the triangle s circumcenter. b. Treat the towns as vertices of a triangle. The center must be placed at the triangle s incenter. c. Treat the towns as sides of a triangle. The center must be placed at the triangle s circumcenter. d. Treat the towns as sides of a triangle. The center must be placed at the triangle s incenter. 2. Find the orthocenter of ABC with vertices A(1, 3), B(2, 7), and C( 2, 3). Ê a. 2, 17 ˆ Ê Ë Á c. 2, 16 ˆ Ë Á Ê b. 2, 13 ˆ Ê Ë Á d. 2, 11 ˆ Ë Á 3. Tell whether a triangle can have sides with lengths 4, 2, and 7. a. No b. Yes 1

2 4. Write a two-column proof. Given: CA is the perpendicular bisector of DB DCA BCA Prove: DC BC Complete the proof. Proof: Statements Reasons 1. F is the midpoint of DB 1. Given DCA BCA 2. DF FB 2. Definition of a bisector 3. [1] 3. Definition of a perpendicular bisector 4. DFC BFC 4. [2]. DC BC. [3] a. [1] m DFC = m BFC = 90 [2] AAS Congruence Theorem [3] Property of congruent triangles b. [1] DF FB [2] Hinge Theorem [3] Comparison Property of Inequality c. [1] DFC BFC [2] Comparison Property of Inequality [3] Converse of the Hinge Theorem d. [1] m DFC = m BFC = 90 [2] SSS Congruence Theorem [3] Comparison Property of Inequality. The size of a TV screen is given by the length of its diagonal. The screen aspect ratio is the ratio of its width to its height. The screen aspect ratio of a standard TV screen is 4:3. What are the width and height of a 27" TV screen? a. width: 21.6 in., height: 16.2 in. c. width: 21.6 in., height:.4 in. b. width: 16.2 in., height: 21.6 in. d. width:.4 in., height: 21.6 in. 2

3 6. Tell if the measures 6, 14, and 13 can be side lengths of a triangle. If so, classify the triangle as acute, right, or obtuse. a. Yes; acute triangle c. Yes; right triangle b. Yes; obtuse triangle d. No. 7. An architect designs the front view of a house with a gable roof that has a triangle shape. The overhangs are 0. meter each from the exterior walls, and the width of the house is 16 meters. What should the side length l of the triangle be? Round your answer to the nearest meter. a. 12 m c. 24 m b. 11 m d. 23 m 8. Tell whether the figure is a polygon. If it is a polygon, name it by the number of its sides. a. polygon, decagon c. polygon, dodecagon b. polygon, hexagon d. not a polygon 9. Polygon ABCDEFGHIJKL is a regular dodecagon (12-sided polygon). Sides EF and GH are extended so that they meet at point O in the exterior of the polygon. Find m FOG. a. m FOG = 100 c. m FOG = 120 b. m FOG = 11 d. m FOG = 110 3

4 10. Show that GHIJ is a parallelogram for x = and y = 8. Complete the explanation. HI = x 10 GJ = 7x 20 Given HI = () 10 = [1] GJ = 7() 20 = [2] Substitute and simplify. GH = 3y JI = y 16 Given GH = 3(8) = [3] JI = (8) 16 = [4] Substitute and simplify. Since HI = GJ and GH = JI, GHIJ is a parallelogram because []. a. [1] 1 [2] 1 [3] 24 [4] 24 [] both sets of opposite sides are congruent. b. [1] 1 [2] 24 [3] 1 [4] 24 [] one set of opposite sides is parallel and congruent. c. [1] 1 [2] 1 [3] 24 [4] 24 [] both sets of opposite sides are parallel. d. [1] 24 [2] 24 [3] 1 [4] 1 [] both sets of opposite angles are congruent. 4

5 11. Show that quadrilateral DEFG is a parallelogram. slope of DE = ( ) = 3 8 slope of FG = = 3 8 DE = (3 ( )) 2 + (10 7) 2 = 73 FG = (8 0) 2 + (4 1) 2 = 73 Complete the explanation. DE and FG have the same slope, so [1]. Since DE = FG, [2]. Because [3], DEFG is a parallelogram. a. [1] DE FG [2] DE FG [3] One pair of opposite sides is equal and perpendicular. b. [1] DE Ä FG [2] DE FG [3] One pair of opposite sides is parallel and congruent. c. [1] DE Ä FG [2] DE ~ FG [3] One pair of opposite sides is parallel and in proportion. d. [1] DE FG [2] DE FG [3] One pair of opposite sides is not congruent but is perpendicular.

6 12. Show that all four sides of square ABCD are congruent and that AB BC. a. AB = 3, BC = 3, CD = 3, DA = 3, slope of AB = 2, slope of BC = 1 2. Since the product of the slopes is 1, AB BC. b. AB = 9, BC = 9, CD = 9, DA = 9, slope of AB = 2, slope of BC = 1. Since the product of the slopes is 1, AB BC. c. AB = 3, BC = 3, CD = 3, DA = 3, slope of AB = 1, slope of BC = 2. 2 Since the product of the slopes is 1, AB BC. d. AB = 4, BC = 4, CD = 4, DA = 4, slope of AB = 2, slope of BC = 1. Since the product of 2 the slopes is 1, AB BC. 6

7 13. Write a two-column proof. Given: ABCD is a rectangle. W, X, Y, and Z are midpoints. Prove: WXYZ is a rhombus. Complete the proof. Proof: Statements Reasons 1. ABCD is a rectangle. 1. Given W, X, Y, and Z are midpoints. 2. A, B, C, and D are right angles 2. Definition of a rectangle 3. A B C D 3. Right Angle Theorem 4. AB CD, BC AD 4. Theorem: Both pairs of opposite sides are congruent.. AW = 1 AD, WD = 1 AD, 2 2. [1] AX = 1 2 AB, XB = 1 2 AB, BY = 1 2 BC, YC = 1 2 BC, CZ = 1 2 CD, ZD = 1 2 CD 6. 1 AB = 1 CD, Division Property of Equality 7. AB = AB, BC = BC 7. Reflexive Property of Equality CD = CD, AD = AD 8. AW = WD = BY = YC, AX = XB = CZ = ZD 9. AW WD BY YC, 8. Substitution 9. Definition of Congruent Segments AX XB CZ ZD 10. AXW BXY DZW CZY 10. [2] 11. WX XY YZ ZW 11. CPCTC 12. WXYZ is a rhombus 12. Definition of a rhombus 7

8 a. [1] Midpoint Theorem [2] SAS b. [1] Definition of Midpoint [2] SAS c. [1] Division Property of Equality [2] ASA d. [1] Division Property of Equality [2] SAS 14. Which of the following is the best name for figure MNOP with vertices M( 3,), N(0,9), O(4,6), and P(1,2)? a. parallelogram c. rhombus b. rectangle d. square 1. Find RS. a. RS = 18 c. RS = 20 b. RS = 24 d. RS = 16 8