H4. The diagram shows a right-angled triangle $A C D$ with a point $B$ on the side $A C$.
The sides of triangle $A B D$ have lengths 3,7 and 8 , as shown.
What is the area of triangle $B C D$ ?
Solution
Let $B C$ equal $b$ and $C D$ equal $h$, as shown in the following diagram.

Using Pythagoras' Theorem in both the triangle $B C D$ and the triangle $A C D$, we get the two equations
$$
\begin{aligned}
b^{2}+h^{2} & =7^{2} \\
\text { and } \quad(b+3)^{2}+h^{2} & =8^{2} .
\end{aligned}
$$

Subtracting equation (1) from equation (2), we obtain
$$
(b+3)^{2}-b^{2}=8^{2}-7^{2} .
$$

Factorising the difference of two squares, we get
$$
(b+3-b)(b+3+b)=(8-7)(8+7)
$$
so that
$$
3(2 b+3)=15 \text {. }
$$

Therefore
$$
b=1 \text {. }
$$

Using equation (1), we now obtain $1+h^{2}=7^{2}$, and so $h=\sqrt{48}=4 \sqrt{3}$.
Therefore the area of the triangle $B C D$ is $\frac{1}{2} \times 1 \times 4 \sqrt{3}$, that is, $2 \sqrt{3}$.