18. The sum of the areas of the squares on the sides of a right-angled isosceles triangle is $72 \mathrm{~cm}^{2}$. What is the area of the triangle?
A $6 \mathrm{~cm}^{2}$
B $8 \mathrm{~cm}^{2}$
C $9 \mathrm{~cm}^{2}$
D $12 \mathrm{~cm}^{2}$
E $18 \mathrm{~cm}^{2}$
18. C Let the length of each of the shorter sides of the triangle be $x \mathrm{~cm}$ and the length of its hypotenuse be $y \mathrm{~cm}$. Then, by Pythagoras' Theorem: $x^{2}+x^{2}=y^{2}$. So $y^{2}=2 x^{2}$. Also, $x^{2}+x^{2}+y^{2}=72$, so $4 x^{2}=72$, that is $x^{2}=18$. Now the area of the triangle, in $\mathrm{cm}^{2}$, is $\frac{1}{2} \times x \times x=\frac{1}{2} x^{2}=9$.