# Honors Geometry Review Exercises for the May Exam

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1 Honors Geometry, Spring Exam Review page 1 Honors Geometry Review Exercises for the May Exam C 1. Given: CA CB < 1 < < 3 < congruent Prove: CAM CBM Proof: 1 A M B 1. < 1 < 1. given. < 1 is supp to < CAM. Linear pairs are supp. 3. < is supp to < CBM 3. Linear pairs are supp. 4. < CAM < CBM 4. Supps of congruent angles are 5. < 3 < 4 5. Given 6. CAM CBM 6. ASA. Given: AB CD AB CD Prove: ABD CDB Proof: D C A B 1. AB CD 1. given. AB CD. Given 3. < ABD <CDB 3. P AIC 4. DB BD 4. Reflexive for 5. ABD CDB 5. SAS 16

3 Honors Geometry, Spring Exam Review page 3 4. AD CB 4. Defn parallelogram 5. Given: BA BC DB bisects < ABC Prove: < AED < CED Proof: 1. BA BC 1. Given. DB bisects < ABC. Given 3. < ABE < CBE 3. Defn bisect 4. BE BE 4. Reflexive for 5. ABE CBE 5. SAS 6. < BEA <BEC 6. CPCTC 7. < BEA is supp to < AED 7. Linear paris are supp. 8. < BEC is supp to < CED 8. Linear paris are supp. 9. < AED < CED 9. Congruent supps. theorem 6. Given: AB CB, BD = BE B Prove: AF CF

4 Honors Geometry, Spring Exam Review page 4 Proof: D E F A C 1. AB CB 1. Given. < B < B. Reflexive for congruence 3. BD = BE 3. Given 4. BD BE 4. Defn congruent 5. ABE CBD 5. SAS 6. < A < C 6. CPCTC 7. AB = CB 7. Defn congruent 8. AD + BD = AB 8. Segment addition postulate 9. CE + BE = BC 9. Segment addition postulate 10. AD + BD = CE + BE 10. Substitution 11. AD = CE 11. Subtraction property of = 1. AD CE 1. Defn congruent 13. < AFD < CFE 13. Vertical angles are congruent 14. AFD CFE 14. AAS 15. AF CF 15. CPCTC

5 Honors Geometry, Spring Exam Review page 5 Solve for the missing parts of each triangle. Give all possible solutions x 63 x 68 Third < = 90 0 Solve for x. Solve or x x = (40)(58)cos15 0 sin 68 0 = 63 x = x = 63 sin 68 x = = A Solve for all angle measures. a = b + c - bc*cosa C 53 B 53 = (0)(180)cosA cosa = -(0)(180) =0.10 m< A = sin B sin( ) sinb = m<b 580. So m<c Solve for all missing lengths and angles. A

6 Honors Geometry, Spring Exam Review page C B BC sin 5 = 19.8 m<b = 38 0 AC cos 5 = 19.8 BC 15.6 AC Solve for all missing parts. A 4 6 B 48 C sinb = sin 48 sin B 6sin 48 4 sin B = m<b = or m< B = So m< A = or m<a = a a = sin 48 sin sin 78.4 sin 48 a = a 3 4 a a = sin 48 sin 5.6 4sin 5.6 sin 48 a = Robin sits at the top of a 35 foot tree looking down at an earthworm crawling along the ground towards the base of the tree. He measures the angle of depression to the worm and finds it to be 54. How far is the earthworm from the base of the tree? tan 54 = 35 e e = 35 tan 54

7 Honors Geometry, Spring Exam Review page 7 e 5.4 ft. 13. A ship is sailing east at 6 knots ( 6 nautical miles per hour) when the captain sightss a small island at an angle of 15 to the north of the ship s course. After 10 minutes, the angle is 8. How far away is the island at the time of the second sighting? 1 6 knots * 6 hours = 1 nautical mi = distance 1 sin13 x sin15 x = sin15 sin13 x = naut. miles m < 1 = 0 m < = 60 m < 3 = 1000 m < 4 = 0 m < 1 = 36 m < = Describe how you would construct the centroidd of a triangle. The centroid of a triangle is a the point where the medians concur. 17. Describe where the orthocenter of an obtuse triangle is.

8 Honors Geometry, Spring Exam Review page 8 The orthocenter is on the outside of a triangle (where the altitudes concur). 18. a) What is the point of concurrence of the medians called? centroid b) If the three medians are constructed in a triangle, and one of the medians has length 18, into what lengths is it divided by the point named in part (a)? 1 and 6 (1 is from the vertex to the centroid, and the 6 is from the centroid to the midpoint of the opposite side.) 19. Calculate ALL of the angles in the figure on the next page. The outside polygon is a regular pentagon. Mark the angles on the drawing Is ACDE a trapezoid? Give evidence for your conclusion. Since m < ACD = 7 and m < EDC = 108, then the same side interior angles are supplementary. Thus AC // DE but AE is not parallel to DC. So ACDE is a trapezoid. D E C F A B 0. Name the largest and smallest angle of the triangle. B 9 1 Largest angle = < C Smallest angle = < A A 3 C

9 Honors Geometry, Spring Exam Review page 9 1. Name the longest side in the triangle A B C m< A = 35 0 Longest side = AB. Complete the chart by checking the figures for which each property is true.

10 Honors Geometry, Spring Exam Review page Given: ABCD DA DE Prove: EDBC is a Proof: 1. Parallelogram ABCD 1. Given. AD BC. Opp sides of parallelograms are congruent 3. AD BC 3. defn parallelogram 4. DA DE 4. Given 5. DE BC 5. Transitive for congruence 6. EDBC is a parallelogram 6. If one set of opp sides of a quad are congruent and parallel, then the quad is a parallelogram 4. Given: MNOP A, B, C, and are the midpoints of MN, NO, OP, and PM respectively Prove: ABCD is a parallelogram Proof: 1. Parallelogram MNOP 1. Given. A, B, C, and are the midpoints. given of MN, NO, OP, and PM respectively 3. Draw MO 3. Through points, there is exactly one line 4. DC and AB are midsegments 4. Def midsegments of MOP and MON 5. DC MO and AB MO 5. Midsegments are parallel to the third side of the triangle

11 Honors Geometry, Spring Exam Review page DC MO and AB MO 6. Midsegments measure one-half of the third side of the triangle 7. DC AB 7. Transitive for 8. DC = AB 8. Transitive for = 9. DC AB 9. Def congruent 10. ABCD is a parallelogram 10. If one set of opp sides of a quad are congruent and parallel, then the quad is a parallelogram 5. A(-4, 0) B(0, 3) C(1,0) D(1, -4) a) Show that ABCD is a trapezoid. m BC = m DA = 1 so BC DA 4 b) Analyze the diagonals of ABCD. The diagonals are not perpendicular (check slopes of AC and BD ), nor are they congruent, using the distance formula. 6. Rhombus PQRS PQ = 13 SQ = 4 PR = TUVW m < 1 = 16 0 m < = 30 0 m < 3 = 64 0

12 Honors Geometry, Spring Exam Review page 1 8. Kite KITE IE = 6 IT = 5 KS = 3 3 KT = m< SKI = 30 0 m < ITS = tan = LM = 10 MF = Area (LMQF) = 1 6(10 18) What does divided proportionally mean? Divided proportionally means that two segments are each dvidied into the same ratio. We often use this with a triangle where one segment is drawn parallel to a side of the triangle. 31. Solve for x and y ( ) x x 8 3 x 10 y ( ) y y

13 Honors Geometry, Spring Exam Review page Solve for each variable 6 x y 0 z 6 y y 0 y = 10 y = 30 6 x x 6 x = 6*6 = 4*39 y = 39 0 z z 6 z = 0*6 =4*130 z = x x 18 4x x 1 x =

14 Honors Geometry, Spring Exam Review page 14 RI IV EB VE RI RI = Find the geometric mean between 5 and x x 19 x = 95 x = Two similar polygons are shown. Find the values of x, y, and z for each. A. B.

15 Honors Geometry, Spring Exam Review page 15 Find the values of x and y y y + 36=48 y = 4 x x = (7)(14) x = Bobbie Mo Smith, 5 8 tall, wants to estimate the height of her favorite tulip tree on Melrose Ave. She asks Ms. Wilson for a mirror from the science department and heads down the Gordon field. She faces the tree and places the mirror between herself and the tree, on the ground, in such a way that she can see the top of the tree in the mirror. She is 4 feet from the mirror and the mirror is 1 feet from the tree. Her eyes are 4 from the top of her head. About how tall is the tree? t t = t

16 Honors Geometry, Spring Exam Review page 16 Find the value of the variable in each example below x 10 x y x 17 y 17 z 1 z 7z 144 z 16, z 9 1 z 7 Find the lengths of all of the sides of the right triangles

17 Honors Geometry, Spring Exam Review page x + (3x) = 0 x + 9x = x = 400 X = 40 X = Find the measure of the numbered angles, or the measure of x (in #56) m<1 = 1 (68 + 1) = m< = 1 (188-56) = m<1 = 66 0

18 Honors Geometry, Spring Exam Review page = 1 (310-x-x) 8 = 310 x x = 8 x = 114 Fill in the blanks with a correct answer. 57. When you solve a triangle numerically that is given by SSA, then it is possible to find 0, 1, or (give a number or numbers) different solutions. 58. When an angle of a triangle is bisected, then there is a proportion of the two parts of the cut side to the ratio of the adjacent sides. This could also be correctly stated as the ratio of one part of the cut side is to its adjacent side as the other part of the cut side is to its adjacent side. 59. The Triangle Inequality says that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. 60. The geometric mean of two numbers is the square root of their product. 61. When the perpendicular bisectors of the three sides of a triangle concur on one side of the triangle, then the triangle is a right_ triangle. 6. In a triangle, then you calculate the length of the short leg by dividing the longest leg by a factor of 3. The Power Theorems for Circles (which is an application of similar triangles) 63. Given: Plane figure with a circle Fill in the result: (EC) ( DC ) = ( AC ) ( BC )

19 Honors Geometry, Spring Exam Review page 19 Defn: The power of C with respect to the circle is the product (EC)(DC), which is the same value as (AC)(BC) 64. Given: Chords SR and QP intersect in a circle at L Fill in the result: (PL) ( LQ ) = (SL ) ( LR ) Defn: The power of L with respect to the circle is the product (PL)(LQ), which is the same value as (SL)(LR) Solve for x x = (1)(4) (x)(15) = ()(1) x = 4.8 or 4 5 x= 88 5

20 Honors Geometry, Spring Exam Review page x = (1)(1) x = (7)(3)(4)(3) Solve for x. What is the power of C? (13)(7) = (x+5)(5) 91 = 5x = 5x x = 66 5 Power of C = (13)(7) which is Solve for RW. What is the power of W? (RW)(4)=(5)(8) RW = 10 Power of W = (5)(8) = 40

21 Honors Geometry, Spring Exam Review page 1 Calculate the area of each figure. Label your answer (with A = ) and write the formula you used in each case A = ( 1 ) hb b = 1 1 1(10 0) A = ( 1 ) hb b = 1 4( 10) = 180 = 4 7. Rhombus with a diagonal 74. of length 10 and a side of length A = 1 dd = 1 (10(4) = An equilateral triangle whose sides measure 1 meters. A= s 3 = = 36 3 sq. meters = A = ( 1 ) hb b = 1 9(6 15) 75. Find the area of the kite KITE. 1 A = 1 dd = 1 (1)(3) =138

22 Honors Geometry, Spring Semester Exam Review page 76. Find the area of the triangle. A= 1 sin bc A= 1 (17)(6.3) sin 58 = Calculate h. A = bh = (10)(16) = 4h h = 0 3

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