In a certain isosceles right triangle, the altitude to the hypotenuse has length $6$.  What is the area of the triangle?
To solve this problem, I need to recall some basic facts about isosceles right triangles and altitudes to hypotenuses. I know that in any right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments that are proportional to the lengths of the legs. So, if I let $x$ be the length of one leg of the isosceles right triangle, then the hypotenuse has length $x\sqrt{2}$, and the altitude divides it into two segments of length $\frac{x\sqrt{2}}{2}$. I also know that the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the two segments it divides the hypotenuse into. So, in this case, the altitude of length $6$ is the geometric mean of $\frac{x\sqrt{2}}{2}$ and $\frac{x\sqrt{2}}{2}$. This means that $6^2 = \frac{x\sqrt{2}}{2} \cdot \frac{x\sqrt{2}}{2}$. Simplifying, I get $36 = \frac{x^2}{2}$. Multiplying both sides by $2$, I get $72 = x^2$. Taking the positive square root of both sides, I get $x = \sqrt{72}$. Now I have the length of one leg of the isosceles right triangle, and I can use it to find the area of the triangle. The area of a triangle is half the product of the base and the height. In this case, the base and the height are both $x$, so the area is $\frac{1}{2}x^2$. Plugging in $x = \sqrt{72}$, I get $\frac{1}{2}(\sqrt{72})^2$. Simplifying, I get $\frac{1}{2}(72)$. Finally, I get the area of the triangle as $36$.

