Geometry: Common Core (15th Edition)

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

Chapter 4 - Congruent Triangles - Common Core Cumulative Standards Review - Selected Response - Page 280: 12


The area of the rectangle is $48$ square centimeters.

Work Step by Step

To find the area of the rectangle, we need to know the dimensions of the rectangle and plug them into the following formula to calculate area: $A = lw$, where $A$ is the area of the rectangle, $l$ is the length of the rectangle, and $w$ is the width of the rectangle. Let us set up some expressions for this problem. Let $w = width$. If the length $l$ is seven more than three times the width, then we can express $l$ as follows: $l = 3w + 7$ We are also given that the perimeter is $38$ cm. Let us use the formula to calculate the perimeter of a rectangle to figure out its dimensions. The perimeter of a rectangle is given by the following formula: $P = 2l + 2w$, where $P$ is the perimeter, $l$ is the length, and $w$ is the width of the rectangle. Let's plug in what we know into this formula: $38 = 2(3w + 7) + 2w$ Let's use the distributive property first: $38 = 6w + 14 + 2w$ Combine like terms on the right side of the equation: $38 = 8w + 14$ Subtract $14$ from each side of the equation to isolate constants on one side of the equation: $8w = 24$ Divide each side by $8$ to solve for $w$: $w = 3$ Now that we have the value for $w$, we can substitute it into the expression for $l$ to find the length of the rectangle: $l = 3w + 7$ Substitute $3$ for $w$: $l = 3(3) + 7$ Multiply first, according to order of operations: $l = 9 + 7$ Add to solve for $l$: $l = 16$ We now have the dimensions of the rectangle, so we can substitute these values into the formula to find the area of the rectangle: $A = lw$ Substitute the values for $l$ and $w$ in the formula: $A = 3(16)$ Multiply to solve for $A$: $A = 48$ The area of the rectangle is $48$ square centimeters.
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