2015 AMC 10A Problem 11

The ratio $4 : 3$ fixes the shape; only the size can change. So work with one convenient rectangle, and then check that the size does not matter.

1. Take a rectangle with length $4$ and width $3$.
2. The diagonal splits it into two right triangles with legs $3$ and $4$. By the Pythagorean theorem the diagonal is
$$d = \sqrt{3^2 + 4^2} = \sqrt{25} = 5.$$
3. The area is $4 \times 3 = 12$.
4. We want $12 = k \times 5^2 = 25k$, so $k = \frac{12}{25}$.
5. Why does another size give the same $k$? Make the sides $4x$ and $3x$. The diagonal becomes $5x$, so $d^2 = 25x^2$, and the area becomes $12x^2$. The ratio area $\div\, d^2$ is still $\frac{12x^2}{25x^2} = \frac{12}{25}$.

The answer is $\boxed{\textbf{(C)}\ \frac{12}{25}}$.

When a shape is fixed and only its size changes, area grows like length squared. So "area $= k \times (\text{length})^2$" has the same $k$ at every size, and you can find it from one easy example. The $3$-$4$-$5$ right triangle is the example that makes this quick.

Let the sides be $4x$ and $3x$. The diagonal is the hypotenuse of a $3$-$4$-$5$ right triangle scaled by $x$, so
$$d = 5x, \qquad x = \frac{d}{5}.$$
The area is
$$4x \cdot 3x = 12x^2 = 12\left(\frac{d}{5}\right)^2 = \frac{12}{25}\,d^2.$$
Hence $k = \frac{12}{25}$.

The answer is $\boxed{\textbf{(C)}\ \frac{12}{25}}$.