Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 360: 17

Answer

36 + 36$\sqrt 3$ $units^{2}$

Work Step by Step

We know in kite diagonals are perpendicular to each other $\angle$ BOA = $\angle$ BOC = 90$^{\circ}$ As we know one diagonal is perpendicular bisector of other OD=OB = 6 Lets take $\triangle$ BOA It is a 45$^{\circ}$ - 45 $^{\circ}$ - 90$^{\circ}$ triangle therefore OA = 6 units $\triangle$ BOC It is a 30$^{\circ}$ - 60 $^{\circ}$ - 90$^{\circ}$ triangle therefore OC = 6 $\sqrt 3$units The area of Kite ABCD = $\frac{1}{2}$* d1*d2 = $\frac{1}{2}$* BD* AC =$\frac{1}{2}$* 12* (OA+OC) =6*(6+6$\sqrt 3$) = 36 + 36$\sqrt 3$ $units^{2}$
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