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?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$
The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - \left( \dfrac{12}{2} \right)^2} =  \sqrt{64} = 8$
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:
$\boxed{\textbf{(D)}\ 32}$