2015 CEMC Gauss (Grade 8) Problems/Problem 6

Problem

A rectangle has a perimeter of $42$ and a width of $4$ .

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What is its length?

$\textbf{(A)}\ 19 \qquad\textbf{(B)}\ 17 \qquad\textbf{(C)}\ 34 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 38$

Solution

The perimeter of a rectangle is equal to $2l + 2w$ , so we have:

$2l + 2w = p$ , where $p$ is the perimeter of the rectangle, $l$ is the rectangle's length, and $w$ is the rectangle's width. However, the problem gives us that $w = 4$ and $p = 42$ , so we have:

$2l + 2 \times 4 = 42$

$2l + 8 = 42$

$2l = 42 - 8 = 34$

$l = \boxed {\textbf {(B) } 17}$

~anabel.disher

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2015 CEMC Gauss (Grade 8) Problems/Problem 6

Problem

Solution