College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 1 - Problems - Page 19: 10


The area of the garden increases by a percentage of 56%.

Work Step by Step

The original area of the garden is $A_0 = \pi r_0^2$ The original radius $r_0$ increases by 25%, so the new radius $r$ is $1.25 ~r_0$ We can find the ratio of new area $A$ of the garden to the original area $A_0$. $ratio = \frac{A}{A_0} = \frac{\pi r^2}{\pi r_0^2} = \frac{\pi (1.25~r_0)^2}{\pi r_0^2} = \frac{(1.25)^2~r_0^2}{ r_0^2} = (1.25)^2 = 1.56$ The area of the garden increases by a factor of 1.56 which is an increase of 56%.
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