jmc

Let the leg length of the isosceles right triangle be $x$, so the hypotenuse of the triangle has length $x\sqrt{2}$. The hypotenuse of the triangle is a side of the square, so the area of the square is $(x\sqrt{2}{)}^2=2x^2$. The area of the triangle is $(x)(x)/2=x^2/2$. So, the area of the pentagon is $$\frac{x^2}{2}+2x^2=\frac{5x^2}{2}.$$Therefore, the fraction of the pentagon's area that is inside the triangle is $$\frac{x^2/2}{5x^2/2}=\frac{x^2}{2}\cdot \frac{2}{5x^2}=\frac{1}{5}=\boxed{20\%}.$$(As an alternate solution, consider drawing the two diagonals of the square. What do you find?)