2010 AMC 10B Problem 7

To use the rectangle we need its area, which is the triangle's area. The triangle has two equal sides, so it splits neatly into two right triangles.

1. Draw the height from the top corner (where the two sides of length $10$ meet) straight down to the side of length $12$. Because the two slanted sides are equal, this height lands exactly in the middle, cutting the $12$ into $6$ and $6$.
2. Each half is a right triangle with long side $10$ and one short side $6$. By the Pythagorean theorem the height is
$$\sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8.$$
(This is the familiar $6$-$8$-$10$ right triangle.)
3. The triangle's area is $\frac12 \cdot \text{base} \cdot \text{height} = \frac12 \cdot 12 \cdot 8 = 48$.
4. The rectangle also has area $48$ and width $4$, so its length is $48 \div 4 = 12$.
5. Its perimeter is $4 + 12 + 4 + 12 = 32$.

The answer is $\boxed{\textbf{(D)}\ 32}$.

A triangle with two equal sides always splits into two matching right triangles when you draw the height to the third side. That gives the height by the Pythagorean theorem, and from there the area. Read the last line of the question carefully: it asks for the perimeter, not the area or the length.

Drop the altitude from the apex of the isosceles triangle to the side of length $12$. It bisects that side into two segments of length $6$, creating two right triangles with hypotenuse $10$ and leg $6$. The altitude is therefore $\sqrt{10^2 - 6^2} = 8$ (a $6$-$8$-$10$ triangle).

Area of the triangle: $\frac{1}{2}\cdot 12 \cdot 8 = 48$.

The rectangle has width $4$ and area $48$, so its length is $48/4 = 12$. Its perimeter is $2(4 + 12) = 32$.

The answer is $\boxed{\textbf{(D)}\ 32}$.