2010 AMC 10B Problem 7

To find the area of the triangle, we can drop the altitude to the side of length $12.$ Then we have a right triangle with one leg $12 \div 2 = 6$ and hypotenuse $10.$ The other leg has length $$\sqrt{10^2 - 6^2} = \sqrt{64} = 8.$$ The area of the triangle is then $$\dfrac{8 \cdot 12}{2} = 48.$$ The length of the rectangle is then $48 \div 4 = 12.$ Its perimeter is $$2(4 + 12) = 2 \cdot 16 = 32.$$ Thus, D is the correct answer.