## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 8 - Section 8.2 - Perimeter and Area of Polygons - Exercises - Page 360: 11

#### Answer

1764$mm^{2}$

#### Work Step by Step

Given, AB = 39 mm, BC =52mm CD =25mm, DA = 60mm By Brahmaguptas formula, the area of cyclic quadrilateral with sides of length a,b,c and d is A = $\sqrt (s-a)(s-b)(s-c)(s-d)$ s = $\frac{a+b+c +d}{2}$ AB =a = 39 mm, BC = b = 52mm CD = c =25mm, DA = d = 60mm s = $\frac{39+52+25 +60}{2}$ =88mm A = $\sqrt (88-39)(88-52)(88-25)(88-60)$ =$\sqrt 49*36*63*28$ =1764$mm^{2}$ Therefore the area of the cyclic quadrilateral ABCD = 1764$mm^{2}$.

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