## Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole

# Chapter 8 - Review Exercises - Page 398: 7a

18

#### Work Step by Step

We first find the height of the trapezoid. It is a leg of the 45-45-90 triangle containing the side of length 6, so we obtain: $h= 6 \times \frac{\sqrt2}{2} = 3\sqrt2$ We also find the lengths of AE and FD, which are equal: $AE=FD = 6 \times \frac{\sqrt2}{2} = 3\sqrt2$ Thus: $A = 1/2(b_1+b_2)(h) = 1/2( 3\sqrt2+ 3\sqrt2)(3\sqrt2)=18$

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