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 362: 56


$A = \frac{ab+cd}{2}$

Work Step by Step

$s = \frac{a+b+c+d}{2}$ It is given in the question that $a^2+b^2 = c^2+d^2$ We can find an expression for the area of the cyclic quadrilateral: $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ $A = \sqrt{(\frac{a+b+c+d}{2}-a)(\frac{a+b+c+d}{2}-b)(\frac{a+b+c+d}{2}-c)(\frac{a+b+c+d}{2}-d)}$ $A = \sqrt{(\frac{b+c+d-a}{2})(\frac{a+c+d-b}{2})(\frac{a+b+d-c}{2})(\frac{a+b+c-d}{2})}$ $A = \sqrt{(\frac{-a^2-b^2+c^2+d^2+2ab+2cd}{4})(\frac{a^2+b^2-c^2-d^2+2ab+2cd}{4})}$ $A = \sqrt{(\frac{0+2ab+2cd}{4})(\frac{0+2ab+2cd}{4})}$ $A = \sqrt{(\frac{2ab+2cd}{4})(\frac{2ab+2cd}{4})}$ $A = \sqrt{(\frac{2ab+2cd}{4})^2}$ $A = \frac{2ab+2cd}{4}$ $A = \frac{ab+cd}{2}$
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.