## Basic College Mathematics (10th Edition)

Published by Pearson

# Chapter 8 - Geometry - 8.6 Circles - 8.6 Exercises - Page 576: 34

#### Answer

The area of the skating rink is $\approx 2973$ ft$^{2}$

#### Work Step by Step

1. Find the area of the circle Let $C =$ area of the circle $C = \pi r^{2}$ $C = \pi (25^{2})$ $C = \pi \times (625)$ $C = 1963.49508$ ft$^{2}$ Given that a quarter of the circle is taken up by the rectangular part of the rink, that quarter part has to be subtracted from the area of the circle. $= (1963.49...) - (1963.49... \times 0.25)$ $= (1963.49...) - (490.8738...)$ ft$^{2}$ $= 1472.6215...$ ft$^{2}$ 2. Find the area of the rectangle Let $R =$ area of the rectangle $R = length \times width$ $R = 60 \times 25$ $R = 1500$ ft$^{2}$ 3. Add up the $\frac{3}{4}$ area of the circle and the area of the rectangle $= 1500 + (1472.6215...)$ $= 2972.621556$ ft$^{2}$ $\approx 2973$ ft$^{2}$

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