Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 360: 34

Answer

As we know Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD Calculate the area of the triangles and used the values in the above equation. Now solve the equation you get the desired result.

Work Step by Step

Given right $\triangle$ ABC We need to prove h = $\frac{ab}{c}$ Let us take a point which intersect Ab and Let AD = x AB= AD + BD c = x + BD BD = c - x Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD The area of right triangle with legs of length a and b given by A = $\frac{1}{2}$ ab area of triangle ABC = $\frac{1}{2}$ ab Area of triangle ADC = $\frac{1}{2}$ xh Area of triangle BCD = $\frac{1}{2}$(c-x)h Substitute the values of Area in Area $\triangle$ ABC = Area $\triangle$ ADC + Area $\triangle$ BCD $\frac{1}{2}$ ab = $\frac{1}{2}$ xh + $\frac{1}{2}$(c-x)h ab = xh + (c-x)h ab = ch h = $\frac{ab}{c}$ Hence Proved
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