Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 5 - Relationships Within Triangles - Common Core Cumulative Standards Review - Selected Response - Page 347: 4

Answer

$G$

Work Step by Step

Let's figure out the area of the square first so that we know what the area of the rectangle must be. The area of a square is the square of its side. Let's write out this formula: $A = s^2$ Let's plug in what we know: $A = 8^2$ Evaluate the exponent to solve for $A$: $A = 64$ square inches We now know that the rectangle also has an area of $64$ square inches. Let's define some variables: Let $w$ = the width of the rectangle Let $4w$ = the length of the rectangle If we multiply the length and the width of a rectangle together, we will get its area. Let's set up this equation to find the value for the width of the rectangle: $64 = 4w(w)$ Multiply the right side of the equation: $4w^2 = 64$ Divide each side of the equation by $4$: $w^2 = 16$ Take the square root of $16$ to solve for $w$: $w = -4$ or $w = 4$ We can get rid of the negative value because a length cannot be negative. The width of the rectangle is $4$ inches. However, we need to find the length of the rectangle. Recall that the length of the rectangle is four times the width of the rectangle. Let's set up the equation: $l = 4w$ Let's plug in our value for $w$: $l = 4(4)$ Multiply to solve: $l = 16$ The length of the rectangle is $16$ inches. This corresponds to option $G$.
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