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 8.2 - Perimeter and Area of Polygons - Exercises - Page 370: 18

Answer

192 $units^{2}$

Work Step by Step

By using pythagoras theorem in right triangle ABD $BD^{2}$ =$AB^{2}$ + $AD^{2}$ $20^{2}$ = $12^{2}$ + $AD^{2}$ $AD^{2}$ = $20^{2}$- $12^{2}$ = 400 - 144= 256 AD = $\sqrt 256$ = 16 We know in kites adjacent sides are equal AB =BC =12 AD = CD = 16 The area of Kite ABCD = Area of $\triangle$ ABD + Area of $\triangle$ BCD = $\frac{1}{2}$*AD*AB + $\frac{1}{2}$*DC*BC = $\frac{1}{2}$*16*12 + $\frac{1}{2}$*16*12 = 96 + 96 = 192 $units^{2}$
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