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2010 AMC 10B Problems/Problem 7

Revision as of 00:40, 26 November 2011 by Gina (talk | contribs)

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:

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

See Also

2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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