16. The sum of the lengths of the three sides of a right-angled triangle is $16 \mathrm{~cm}$. The sum of the squares of the lengths of the three sides of the triangle is $98 \mathrm{~cm}^{2}$.
What is the area, in $\mathrm{cm}^{2}$, of the triangle?
A 8
B 10
C 12
D 14
E 16
16. A Let the lengths, in $\mathrm{cm}$, of the sides of the right-angled triangle be $p, q$ and $r$, where $r$ is the length of the hypotenuse. The area of the right-angled triangle is $\frac{1}{2} p q$.
We are told that $p+q+r=16$ and $p^{2}+q^{2}+r^{2}=98$.
By Pythagoras' Theorem, $p^{2}+q^{2}=r^{2}$ so $2 r^{2}=98$. Hence $r=7$.
Therefore $p+q=16-7=9$. So $(p+q)^{2}-\left(p^{2}+q^{2}\right)=9^{2}-49=81-49=32$.
Hence $p^{2}+2 p q+q^{2}-p^{2}-q^{2}=32$, so $p q=16$.
Therefore the area of the right-angled triangle is $\frac{1}{2} p q=\frac{1}{2} \times 16=8$.