Elementary Geometry for College Students (7th Edition)

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 8 - Review Exercises - Page 395: 7c


$33 \sqrt3$

Work Step by Step

We first find the height of the trapezoid. It is a leg of the 30-60-90 triangle containing the side of length 6, so we obtain: $h= 6 \times \frac{\sqrt3}{2} = 3\sqrt3$ We also find the lengths of AE and FD, which are equal: $AE=FD = 6 \times \frac{1}{2} = 3$ Thus: $A = 1/2(b_1+b_2)(h) = 1/2(8+14)(3\sqrt3)=33\sqrt3$
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