Chapter 7 & 8 Review Question Answers

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Alberta Ferguson
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1 1. If the midpoints of consecutive sides of a kite are joined in order, what is the most descriptive name of the figure formed? Rectangle If the midpoints of consecutive sides of a rhombus are joined in order, what is the most descriptive name of the figure formed? Rectangle If the midpoints of consecutive sides of a rectangle are joined in order, what is the most descriptive name of the figure formed? Rhombus 2. The sum of the measures of the s of a polgon is Find the number of diagonals for the polgon. ( n - 2)180 = 5580 ( n - 2) = 31 n = 33 n( n - 3) 2 = 33 ( 30 ) 2 = 95 diagonals 3. polgon has 5 diagonals. What is its name? n( n - 3) 2 = 5 n( n-3) = 108 n 2-3n = 0 ( n - ) ( n + 9) = 0 n = or n = -9 ( which can't be since ou can't have a negative number of sides) this is a dodecagon!! Barood Page 1 of 10

2 . The measure of an of an equiangular polgon exceeds four times the measure of one of the polgon's exterior s b 30. What is the name of the polgon? x = interior x = exterior x = ( x) + 30 x = x x = 750 x = 150 Now, an exterior would measure = 30. Since the sum of all the exterior s is 360, the number of s is =. polgon with s is an equiangular dodecagon. 5. What is the name of an equiangular polgon if the ratio of the measure of an interior to the measure of an exterior is 7:2? 7x + 2x = 180 ( since the exterior and interior s must be supplementar) 9x = 180 x = 20 So, an exterior is 2x or 2( 20) = 0. Since the sum of all the exterior s is 360, the number of s is = 9. polgon with 9 s is an equiangular nonagon. Barood Page 2 of 10

3 6. 7 of the s of a decagon have measures whose sum is 20. Of the remaining 3 s, exactl 2 are complementar and exactl 2 are supplementar. Find the measure of the smallest of these = 220 for the three remaining angles x + ( 90 - x) + ( x) = x = 220 x = 50 ( 90 - x) = 0 ( x) = 130 the smallest is 0!! 7. The ratio of the measures of the s of a heptagon is 3::::5:5:5. Find the smallest. 3x + x + x + x + 5x + 5x + 5x = ( 7-2)180 30x = 900 x = 30 the smallest would be 3( 30) = If one exterior angle of a triangle is 110 and one interior angle is 80, the measure of the smallest angle in the triangle is what? 80 The smallest angle measures The measures of the angles of an undecagon sum to what? ( 11-2)180 = ( 9)180 = 1620 Barood Page 3 of 10

4 10. The measure of each exterior angle of a regular nonagon is what? = If 10 + x 15 + = 2 3, find x. 3( 10 + x) = 2( 15 + ) x = x = 2 3x = 2 x = 2 3. Find the mean proportional ( geometric mean) of the numbers 3 & x = x 27 x 2 = 81 x = 81 = ±9 13. Find the 2nd proportional for the numbers 8, 3, and x = x = 33 x = 33 8 Barood Page of 10

5 1. Find x. Explain our reasoning. 9 D Since D C = 9 = 3 CB DC b SS and C CB = 16 = 3 2 x we can set up the following proportion: x B 16 C 2 = 3 x = 72 x = Find x. 10 = x = x 15 x 5x = x = The perimeter of BC = 25. Find BC. B 15 - x = x x x x = 90-6x 10x = 90 x = 9 = BC D 6 C Barood Page 5 of 10

6 17. Find B & BE. 6 BE = = 6 B 6 D 6 BE = = 6 B 6 9 3( BE) = 2 BE = 8 3( B) = 30 B = 10 B E C 18. Find NR + NS. N 5x - 21 x 5x - 21 x = 5 8 R 5 8 S 5x = 0x x = 168 x = = 2 5 NR = 5( 2 5 ) = = 8 NS = NR + NS = = = 10 5 = = 6 5 Barood Page 6 of 10

7 19. Find P HFJ F x - 2 = 9 x + 3 ( x - 2) ( x + 3) = 36 x - 2 x 2 + x - 6 = 36 x 2 + x - 2 = 0 9 G 5 K x + 3 ( x - 6) ( x + 7) = 0 x = 6 and/or x = -7 since FG can't be < 0 H J FG = 6-2 = & KJ = = 9 13 = 5 = 65 = 65 So, P HFJ = = = = Find all possible values for. 27 = 2 = = ±18 However, since can't be < 0, the onl viable answer is = 18. Barood Page 7 of 10

8 21. Find. Using Theorem 66, we can see that = 18 = = 2 = Given: BDF is a! C B Prove: CBD DFE D F Statements E Reasons 1. BDF is a! 1. Given 2. C DF; BD E 2. Opposite Sides of a! are 3. C EDF. CDB E 5. CBD DFE PC PC ( 3, ) Barood Page 8 of 10

9 23. Given: DH HF CB BG C D Prove: CD DE = GF FE Statements B G H F Reasons E 1. DH CB; HF BG 2. CD DE = BH HE 3. GF FE = BH HE. CD DE = GF FE 1. Given Side-Splitter Theorem S.. 2 Subsititution ( 3 into 2) 2. J Given: HR GM Prove: PR OM = PH OG M R O P H G Statements Reasons 1. HR GM 2. JRP M 3. JPR JOM. JPR JOM 5. PR OM = JP JO 6. JPH JOG 6. PC 7. JHP G 7. PC 8. JPH JOG 9. PH OG = JP JO 10. PR OM = PH OG 1. Given PC PC ( 2, 3) CSSTP ( 6, 7) CSSTP 10. Substitution ( 9 into 5) Barood Page 9 of 10

10 25. Given: YSTW is a parallelogram SX YW SV WT Y X W V Prove: SX YW = SV WT S T Statements Reasons 1. YSTW is a! 2. Y T 3. SX YW. SXY is a right 5. SV WT 5. Given 6. SVT is a right 6. Defn. of 7. SXY SVT 7. RT 8. SXY SVT 9. SX SV = SY ST 1. Given 2. Opp. s of! are CSSTP 10. SX ST = SV SY 10. Means-Extremes Products Thm. 11. ST YW; SY WT 11. Opp. sides of parallelogram are. ST = YW; SY = WT. Definition of Segments 13. SX YW = SV WT 13. Substitution ( in 10) Given Defn. of ( 2, 7) Barood Page 10 of 10

9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC.

9. AD = 7; By the Parallelogram Opposite Sides Theorem (Thm. 7.3), AD = BC. 10. AE = 7; By the Parallelogram Diagonals Theorem (Thm. 7.6), AE = EC. 3. Sample answer: Solve 5x = 3x + 1; opposite sides of a parallelogram are congruent; es; You could start b setting the two parts of either diagonal equal to each other b the Parallelogram Diagonals Theorem