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: 18

Answer

A = $\frac{100 \pi}{9} - 25 \frac{\sqrt{3}}{3}$ $ P = =10 + \frac{20\pi}{3\sqrt3} $

Work Step by Step

We first find the radius: $\frac{r}{5} = \frac{2}{\sqrt3} \\ r = \frac{10}{\sqrt3}$ We know that 1/3 of the circumference plus AB is the perimeter. Thus: $ P = 10 + \frac{2\pi(10)}{3\sqrt3} =10 + \frac{20\pi}{3\sqrt3} $ We find the area. To do so, we first must find the distance between the center of the circle and the line segment (the height of the triangle): $ h = \sqrt{( \frac{10}{\sqrt3})^2 - 5^2} = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}$ Thus, we subtract the area of the triangle from 1/3 of the area of the circle: $ A = \pi(1/3) (\frac{10}{\sqrt3})^2 - 1/2(10)(\frac{5}{\sqrt{3}}) = \frac{100 \pi}{9} - 25 \frac{\sqrt{3}}{3}$
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