## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 351: 47

#### Answer

4$\frac{8}{13}$ in

#### Work Step by Step

Given the triangle is right angles triangle and two sides are 5 in and 12 in By using pythagoras theorem $AC^{2}$ =$AB^{2}$ + $BC^{2}$ $AC^{2}$ = $5^{2}$ + $12^{2}$ $AC^{2}$ = 25 + 144 $AC^{2}$ = 169 AC = 13 in Let us assume AD =x AC = AD + DC 13 in = x + DC DC = 13 - x Now apply Pythagoras theorem on triangle BDC $BC^{2}$ =$BD^{2}$ + $DC^{2}$ $12^{2}$ = $BD^{2}$ + $(13 - x)^{2}$ $BD^{2}$ = $12^{2}$ - $(13- x)^{2}$ - Eq 1 Similarly apply Pythagoras theorem on triangle ABD $AB^{2}$ = $AD^{2}$ + $BD^{2}$ $5^{2}$= $BD^{2}$ + $x^{2}$ $BD^{2}$ = $5^{2}$- $x^{2}$ - eq2 $BD^{2}$ is common in both the equations So, Eq1 = eq2 $12^{2}$ - $(13- x)^{2}$ = $5^{2}$- $x^{2}$ 144 - 169- $x^{2}$ + 26x= 25 - $x^{2}$ -25 + 26x = 25 26x = 50 x = $\frac{50}{26}$ Now put the value of x in eq1 $BD^{2}$ = 25 - $\frac{50}{26}^{2}$ $BD^{2}$ = 25 - $\frac{2500}{676}$ $BD^{2}$ = $\frac{16900 - 2500}{676}$ BD = $\sqrt \frac{14400}{676}$ BD = $\frac{120}{26}$ in = 4$\frac{8}{13}$ in

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