jmc

If one acute angle of a right triangle is $45^{\circ }$, then the other is $90^{\circ }-45^{\circ }=45^{\circ }$, so the triangle is a 45-45-90 triangle. Here are two solutions:

Solution 1: Find the legs. The hypotenuse is $\sqrt{2}$ times the length of each leg, so each leg has length $10/\sqrt{2}$. Therefore, the area of the triangle is $$\frac{1}{2}\cdot \frac{10}{\sqrt{2}}\cdot \frac{10}{\sqrt{2}}=\frac{10\cdot 10}{2\sqrt{2}\cdot \sqrt{2}}=\frac{100}{4}=\boxed{25}.$$Solution 2: Find the altitude to the hypotenuse. Altitude $\overline{AD}$ to the hypotenuse of isosceles right triangle $ABC$ below divides $ABC$ into 45-45-90 triangles $ABD$ and $ACD$. Therefore, $AD=BD=CD$, so $D$ is the midpoint of the hypotenuse. This gives us $BD=CD=BC/2=5$, so $AD=5$ and $$[ABC]=\frac{(AD)(BC)}{2}=\frac{(5)(10)}{2}=\boxed{25}.$$