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: 39a

Answer

12 in

Work Step by Step

Lets assume AD = x, So AC = AD+ DC 14 = x +DC DC = 14 - x Now apply Pythagoras theorem on triangle ABD $AB^{2}$ = $AD^{2}$ + $BD^{2}$ $13^{2}$ = $x^{2}$ + $BD^{2}$ $BD^{2}$ = $13^{2}$ - $x^{2}$ - Eq 1 Similarly apply Pythagoras theorem on triangle BCD $BC^{2}$ = $CD^{2}$ + $BD^{2}$ $15^{2}$ = $(14 - x)^{2}$ + $BD^{2}$ $BD^{2}$ = $15^{2}$ - $(14 - x)^{2}$ - eq2 $BD^{2}$ is common in both the equations So, Eq1 = eq2 $13^{2}$ - $x^{2}$ = $15^{2}$ - $(14 - x)^{2}$ 169 - $x^{2}$ = 225 -196 - $x^{2}$ + 28x 169 = 29 + 28x 28x = 169 - 29 28x = 140 x = 5 Now put the value of x in eq1 $BD^{2}$ = $13^{2}$ - $5^{2}$ $BD^{2}$ = 169 - 25 $BD^{2}$ = 144 BD = 12 in
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