Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 2 - Solving Equations - Concept Byte - Finding Perimeter, Area, and Volume - Page 116: 3


$P=26.8in$ $A=52.3in^2$

Work Step by Step

The perimeter of the figure $P$ can be found by adding up all of the exterior side lengths. The perimeter or circumference of the half circle is found with the formula $P_{h}=$$\pi$$r$. Substitute the given value for $r$ and use $3.14$ for $\pi$ in the formula above. $P_{h}$=$(3.14)(3)$ $P_{h}$=$9.42$ Since there are 2 half circles in the figure, simply multiply the answer above by 2: $9.42$$\times$$2$=$18.84$ The exterior side lengths of the rectangle are given as $4$ and $4$ which add up to $8$. Therefore, $P=18.84+8=26.84$ which rounds to $26.8$ The area of the figure $A$ is the sum of the area of the first half circle $A_{h_1}$, the area of the second half circle $A_{h_2}$, and the area of the rectangle $A_{r}$. Use $3.14$ for $\pi$ $A_{h_1}$=$\frac{1}{2}$$\pi$$r^2$ $A_{h_1}$=$\frac{1}{2}$$(3.14)(3^2)$ $A_{h_1}$=$14.13$ $A_{h_2}$=$\frac{1}{2}$$\pi$$r^2$ $A_{h_2}$=$\frac{1}{2}$$(3.14)(3^2)$ $A_{h_2}$=$14.13$ $A_{r}$=$lw$ $A_{r}$=$4(6)$ $A_{r}$=$24$ Add up all of the solved areas to find $A$: $A=14.13+14.13+24=52.26$ which rounds to $52.3$
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