2010 AMC 10B Problems/Problem 7

Problem

A triangle has side lengths $10$ , $10$ , and $12$ . A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$

Solution

The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - ( \dfrac{12}{2})^2} = \sqrt{64} = 8$

Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$

We have that the area of the rectangle is the same as the area of the triangle, namely $48$ . We also have the width of the rectangle: $4$ .

The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$

The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$

The answer is:

$\boxed{\textbf{(D)}\ 32}$

An alternative way to find the area of the triangle is by using Heron's formula, $A=\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$ . Thus the area equals $\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48.$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2010 AMC 10B Problems/Problem 7

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