4.10 Proportional Segments in Right Triangles Theorem 61 It in a right triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse: 1. The two triangles thus formed are similar to the given tri- angle and to each other. II. The perpendicular is the mean proportional between the segments of the hypotenuse. II. Each leg of the given triangle is the mean proportional between the hypotenuse and the adjacent segment of the hypotenuse. Given: Right AABC, and CD the I from C to the hypotenuse AB Prove: I. AACD ~ AABC ~ ACBD II. AD : CD = CD : DB – AD = CD Bb BC BA %3D B. AC DB ep вс III. AD : AC = AC : AB and BD : BC = BC : BA ло %3D AB %3D S7. If BC = 8 in., AD = 12 in., find DB, AC, BC %3D %3D and DC. Suggestion: Let x = DB and use Theo- rem 61 III. %3D

4.10 Proportional Segments in Right Triangles Theorem 61 It in a right triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse: 1. The two triangles thus formed are similar to the given tri- angle and to each other. II. The perpendicular is the mean proportional between the segments of the hypotenuse. II. Each leg of the given triangle is the mean proportional between the hypotenuse and the adjacent segment of the hypotenuse. Given: Right AABC, and CD the I from C to the hypotenuse AB Prove: I. AACD ~ AABC ~ ACBD II. AD : CD = CD : DB – AD = CD Bb BC BA %3D B. AC DB ep вс III. AD : AC = AC : AB and BD : BC = BC : BA ло %3D AB %3D S7. If BC = 8 in., AD = 12 in., find DB, AC, BC %3D %3D and DC. Suggestion: Let x = DB and use Theo- rem 61 III. %3D

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.1: Measures Of Center

Problem 14PPS

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