Solution
A
We let $S$ be the foot of the perpendicular from the point $I$ to $H J$. We let $h, x$ and $y$ be the lengths of $I S, \quad H S$ and $S J$, respectively.

The square GQOP has area 10 and hence side length $\sqrt{10}$.

Similarly, the side length of the square $H J N O$ is $\sqrt{90}=3 \sqrt{10}$ and the side length of the square $R K M N$ is $\sqrt{40}=2 \sqrt{10}$.
We have $G Q=\sqrt{10}$ and $H Q=H O-Q O=3 \sqrt{10}-\sqrt{10}=2 \sqrt{10}$. We leave it to the reader to check that the triangles $G P F, H Q G$ and $I S H$ are similar. It follows that $F P=\frac{1}{2} \sqrt{10}$ and $h=2 x$. We also have $J R=J N-R N=3 \sqrt{10}-2 \sqrt{10}=\sqrt{10}$ and $R K=2 \sqrt{10}$. The triangles $K M L$, $J R K$ and $I S J$ are similar. Therefore $M L=4 \sqrt{10}$ and $h=\frac{1}{2} y$.
Since $h=2 x$ and $h=\frac{1}{2} y$, it follows that $y=4 x$. Also $x+y=3 \sqrt{10}$. Therefore $5 x=3 \sqrt{10}$. Hence $x=\frac{3}{5} \sqrt{10}, y=\frac{12}{5} \sqrt{10}$ and $h=\frac{6}{5} \sqrt{10}$.
We now have that the base of the triangle $F I L$ is
$$
F L=F P+P O+O N+N M+M L=\frac{1}{2} \sqrt{10}+\sqrt{10}+3 \sqrt{10}+2 \sqrt{10}+4 \sqrt{10}=\frac{21}{2} \sqrt{10} .
$$
and the height of the triangle is
$$
I S+H O=\frac{6}{5} \sqrt{10}+3 \sqrt{10}=\frac{21}{5} \sqrt{10} .
$$

Hence the area of the triangle $F I L$ is given by
$$
\frac{1}{2}(\text { base } \times \text { height })=\frac{1}{2}\left(\frac{21}{2} \sqrt{10} \times \frac{21}{5} \sqrt{10}\right)=\frac{1}{2}\left(\frac{21^{2}}{10} \times 10\right)=220.5 \text {. }
$$

For INVESTIGATION
23.1 Prove (a) that the triangles $G P F, H Q G$ and $I S H$ are similar, and
(b) that the triangles $K M L, J R K$ and $I S J$ are similar.
23.2 Show that the side $F I$ of the triangle $F I L$ is perpendicular to the side $L I$.
23. Three squares $G Q O P, H J N O$ and $R K M N$ have vertices which sit on the sides of triangle FIL as shown. The squares have areas of 10 , 90 and 40 respectively.
What is the area of triangle $F I L$ ?
A 220.5
B $\frac{21}{5} \sqrt{10}$
C 252
D $\frac{21}{2} \sqrt{10}$
E 441