6. A right triangle has a hypotenuse of length 2, and one of its legs has length 1. The altitude to its hypotenuse is drawn. What is the area of the rectangle whose diagonal is this altitude?
Solution: Call the triangle $A B C$, with $A C=2$ and $B C=1$. By the Pythagorean theorem, $A B=\sqrt{3}$. Call the point at which the altitude intersects the hypotenuse $D$. Let $E \neq B$ be the vertex of the rectangle on $A B$ and $F \neq B$ be the vertex of the rectangle on $B C$. Triangle $B D C$ is similar to triangle $A B C$, so $B D=\frac{\sqrt{3}}{2}$. Triangle $D B F$ is similar to triangle $A B C$, so $D F=\frac{\sqrt{3}}{4}$ and $B F=\frac{3}{4}$. The area of the rectangle is thus $\frac{\sqrt{3}}{4} \frac{3}{4}=\frac{3 \sqrt{3}}{16}$.