14. A square $A B C D$ has side 40 units. Point $F$ is the midpoint of side $A D$. Point $G$ lies on $C F$ such that $3 C G=2 G F$.
What is the area of triangle $B C G$ ?
Solution
320
Start by drawing a diagram.
By Pythagoras' Theorem, $F C=20 \sqrt{5}$ and hence $C G=8 \sqrt{5}$. Now, $\sin \angle B C G=\sin \angle C F D=$ $\frac{40}{20 \sqrt{5}}=\frac{2}{\sqrt{5}}$.
Hence the area of triangle $B C G$ is $\frac{1}{2} \times 40 \times 8 \sqrt{5} \times \frac{2}{\sqrt{5}}=320$.