1 7.1 Triangle Application Theorems (pg )
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1 Geometry for Enjoyment and Challenge - Text Solutions Ruth Doherty Triangle Application Theorems (pg ) 2. Given: m 1 = 130 m 7 = Prove: Find the measures of 2, 3, 4, 5, m 1 + m 2 = 180 (Definition of Supplementary) m 2 = 50 m 7 + m 6 = 180 (Definition of Supplementary) m 6 = 110 m 7 = m 5 (Vertical Angles are congruent, so their measures are equal) m 5 = 70 m 2 + m 3 + m 5 = 180 (Sum of the measures of the interior angles of a triangle) m = 180 m 3 = 60 m 4 = m 5 + m 2 (Exterior Angle Equality) m 4 = 120
2 3. C Given: m CAB = 80 m CBA = 60 AE and BD are altitudes Find: m C and m AF B E D F A B m CAB + m CBA + m C = 180 (Sum of the measures of the interior angles of a triangle) m C = 180 m C = 40 m ADB = 90 (Altitudes are perpendicular to the side of the triangle,, = 90 ) m BEA = 90 m DBA + m ADB + m CAB = 180 (Sum of the measures of the interior angles of a m DBA = 180 triangle) m DBA = 10 m EAB + m AEB + m CBA = 180 (Sum of the measures of the interior angles of a m DBA = 180 triangle) m DBA = 30 m AF B = 180 m DBA m EAB (Sum of the measures of the interior angles of a ) m AF B = 140
3 4. G In the diagram as marked, if m G = 50, find m M M H J Let m MHJ = x and m MJH = y, so m MHJ + m MJH = x + y 2x + 2y + 50 = 180 (Sum of the measures of the interior angles of a triangle) 2(x + y) = 130 x + y = 65 m M +m MHJ +m MJH = 180 (Sum of the measures of the interior angles of a triangle) m M = 180 (x + y) m M = m M = 115
4 5. The measure of the three angles of a triangle are in the ratio of 4 : 5 : 6. Find the measure of each. Let the measures of the three angles be x, y, and z. So because the sum of the measures of the interior angles is 180, x + y + z = 180. Now, x = 4s, y = 5s, z = 6s because we want the angles to be in a ratio of 4 : 5 : 6. 4s + 5s + 6s = 180 by substitution. 15s = 180 s = 12 x = 48, y = 60 and z = 72
5 6. O Given: m ORS = 4x + 6 m P = x + 24 m O = 2x + 4 Find: m O S R P By the Exterior Angle Equality: m ORS = m O + m P 4x + 6 = (x + 24) + (2x + 4) 4x + 6 = 3x + 28 x = 22 m O = 2(22) + 4 = 48
6 9. Tell whether each statement is true Always, Sometimes or Never. (a) The acute angles of a right triangle are complementary. Always (b) The supplement of one of the angles of a triangle is equal in measure to the sum of the other two angles of the triangle. Always (c) A triangle contains two obtuse angles. Never (d) If one of the angles of an isosceles triangle is 60, the triangle is equilateral. Always (e) If the sides of one triangle are doubled to form another triangle, each angle of the second triangle is twice as large as the corresponding angle of the first triangle. Never
7 12. In DEF, the sum of the measures of D and E is 110. The sum of the measures of E and F is 150. Find the sum of the measures of D and F. A D 150 E F 110 C By the Exterior Angle Equality, we know: E + EDF = ADF and E + DF E = DF C. So, m DF C = 110 and m ADF = 150, then because DF C$ EDF, m EDF = 30. DF C$ EF D as well, so m EF D = 70. Thus, m D + m F = 100
8 17. Given: EF GH is a rectangle. F H = 20 J, K, M, and O are midpooints a) Find the perimeter of JKMO b) What is the most descriptive name for JKMO? E K F J M H O G (a) J is the midpoint of EH and K is the midpoint of EF. Then by the midline theorem, KJ F H and KJ = 1F H. By the same reasoning, OM F H and MO = 1 F H. So 2 2 KJ = 10 and MO = 10. EF GH is a rectangle so the diagonals are congruent, which means EG = 20. By the midline theorem, JO EG and JO = 1 EG, KM EG and 2 KM = 1 EG. Therefore KM = 10 and JO = 10. So the perimeter of KJMO is (b) The sides of JKMO are all congruent. The opposite sides are also parallel, so JKMO is a rhombus.
9 2 7.2 Two Proof-Oriented Triangle Theorems (p ) 13. Given: OHJM is an isosceles trapezoid, with bases HJ and OM. HP J = JKH Prove: (a) HRJ is isosceles (b) HP = JK P O R M K (c) R is equidistant from O and M. H J (a) It is given that HP J = JKH. OHJM is an isosceles trapezoid which means that the bases angles are congruent, P HJ = KJH. By the reflexive property, HJ = HJ. So JHP = HJK by AAS. Then KHJ = P JH by CPCTC. Therefore, by, HRJ is isosceles. (b) Also by CPCTC, P H = KJ. (c) HRJ is isosceles so RH = RJ. And by CPCTC, HK = P J. Therefore by the subtraction property, P R = RK. Because OHJM is an isosceles trapezoid, the legs are congruent so OH = MJ. Thus by the subtraction property, OP = MK. HP R$ OP R and JKR$ MKR, because they form straight angles. Given HP R = JKR, by the congruent supplement theorem, OP R = MKR. Then OP R = MKR by SAS. Which means OR = MR by CPCTC.
10 14. Z Given: AC XY AB CY ZAC = XAB Prove: X = Z A C X B Y Statements Reasons 1. AC XY 1. Given 2. AB CY 2. Given 3. ZAC = XAB 3. Given 4. CY B = ABX 4. P CAC 5. CY B = ZCA 5. P CAC 6. CY B = ZCA 6. Transitive Property 7. X = ZAC 7. P CAC 8. X = XAB 8. Transitive Property 9. Z = XAB 9. No Choice Theorem 10. X = Z 10. Transitive Property
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