Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 8 - Applications of Trigonometric Functions - 8.4 Area of a Triangle - 8.4 Assess Your Understanding - Page 544: 39


$216.5$ square inches

Work Step by Step

The top triangle has sides: $12,12, 17.$ Hence, by Heron's formula, $K=\sqrt{s(s-a)(s-b)(s-c)}$, where $s=\frac{a+b+c}{2},$ we have: $s=\frac{12+12+17}{2}=20.5$ $K=\sqrt{20.5(20.5-12)(20.5-12)(20.5-17)}\approx72$ square inches The bottom rectangle has sides, $17, 8.5, 17, 8.5$, hence its area is $17\cdot8.5=144.5$. Thus, the total area is $=72+144.5=216.5$ square inches
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