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: 40a

Answer

8 cm

Work Step by Step

Lets assume AD = x, So AC = AD+ DC 21 = x +DC DC = 21 - x Now apply Pythagoras theorem on triangle ABD $AB^{2}$ =$AD^{2}$ + $BD^{2}$ $10^{2}$ = $^{2}$ + $BD^{2}$ $BD^{2}$ = $10^{2}$- $x^{2}$ - Eq 1 Similarly apply Pythagoras theorem on triangle BCD $BC^{2}$ = $CD^{2}$ + $BD^{2}$ $17^{2}$= $(21-x)^{2}$ + $BD^{2}$ $BD^{2}$ = $17^{2}$- $(21-x)^{2}$ - eq2 $BD^{2}$ is common in both the equations So, Eq1 = eq2 $10^{2}$ - $x^{2}$ = $17^{2}$ - $(21-x)^{2}$ 100 - $x^{2}$ = 289 -441 - $x^{2}$+ 42x 100 = -152 + 42x 42x = 252 x = 6 Now put the value of x in eq1 $BD^{2}$ = 100 - 36 $BD^{2}$ = 64 BD = 8 cm
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