Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 6 - The Circular Functions and Their Graphs - 6.1 Radian Measures - 6.1 Exercises: 110

Answer

The new area is twice the original area . In general, this result holds for any values of $\theta$ and $r$. (refer to the explanation below)

Work Step by Step

RECALL: The area of a sector $(A)$ intercepted by a central angle $\theta$ on a circle whose radius is $r$ is given by the formula: $A = \frac{1}{2}r^2\theta$, where $\theta$ is in radian measure. From Number 109, the area of the sector is $\color{blue}{A \approx 8,060 \space yd^2}$. If the angle is halved ($\frac{\pi}{9}$ and the radius length is doubled ($304 \text{yd}$), the area of the sector becomes: $A=\frac{1}{2}r^2\theta \\A=\frac{1}{2}(304^2)(\frac{\pi}{9}) \\\color{blue}{A \approx 16,130 \space yd^2}$ This is twice the area of the sector in Number 109. This result hold in general for any values of $\theta$ and $r$ because if the radius is doubled and the angle is halved, then: Area of Original Sector: $A = \frac{1}{2}r^2\theta$ Area of New Sector: $A=\frac{1}{2}(2r)^2\cdot \frac{\theta}{2}=\frac{1}{2}(4r^2)(\frac{\theta}{2})=r^2\theta$ Notice that the area of the new sector is twice the area of the original sector.
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