Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 392: 19

Answer

( 25$\sqrt 3$ - $\frac{25 \pi}{2}$) $cm^{2}$

Work Step by Step

To find the area A of the shaded region Area of triangle = Area of shaded region +3 * Area of sector Area of shaded region = Area of triangle - 3 * area of sector As we know area of equilateral triangle = $\frac{\sqrt 3}{4} side^{2}$ Area of Sector = $\frac{m}{360}$ * $\pi r^{2}$ = $10^{2} \frac{\sqrt 3}{4}$ - 3 * $\frac{60}{360} \pi 5^{2}$ =$100 \frac{\sqrt 3}{4}$ - 3 * $\frac{1}{6} \pi * 25$ =( 25$\sqrt 3$ - $\frac{25 \pi}{2}$) $cm^{2}$
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