2010 AMC 10B Problems/Problem 7

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 therefor is: $l = \dfrac{48}{4} = 12$

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

The answer is:

$\boxed{\mathrm{(D)} = 32}$

Page