Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 360: 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 $units^{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.