33. The figure below shows a trapezium $A B C D$ in which $A D / / B C$ and $B C=3 A D . F$ is the midpoint of $A B$ and $E$ lies on $B C$ extended so that $B C=3 C E$. The line segments $E F$ and $C D$ meet at the point $G$. It is given that the area of triangle $G C E$ is $15 \mathrm{~cm}^{2}$ and the area of trapezium $A B C D$ is $k \mathrm{~cm}^{2}$. Find the value of $k$.
33. Answer: 360

Extend $B A$ and $C D$ to meet at $X$. Let $H$ be the point on $C D$ such that $F H / / B C$. Let $A D=C E=a$. Then $B C=3 a$, and $F H=\frac{1}{2}(A D+B C)=2 a$.
By the similarity of triangles $F H G$ and $E C G$, we have
(i) area of $\triangle F H G=\left(\frac{F H}{C E}\right)^{2} \times$ area of $\triangle E C G=60 \mathrm{~cm}^{2}$;
(ii) $\frac{H G}{C G}=\frac{F H}{C E}=2$, so that $H G=2 C G$ and $D H=H C=\frac{3}{2} H G$.

It follows from (i) and (ii) that the area of triangle $F D H=\frac{3}{2} \times 60=90 \mathrm{~cm}^{2}$. Now, let area of triangle $X A D$ be $y \mathrm{~cm}^{2}$. By the similarity of triangles $X A D$ and $X F H$, we have
(iii) $\frac{X A}{X F}=\frac{A D}{F H}=\frac{1}{2}$, so that $X A=A F$ and hence area of $\triangle X D F=2 \times$ area of $\triangle X A D=2 \mathrm{y} \mathrm{cm}^{2}$;
(iv) area of $\triangle X F H=\left(\frac{F H}{A D}\right)^{2} y=4 y$.

It follows from (iii) and (iv) that the area of triangle $F D H=4 y-2 y=2 \mathrm{ycm}^{2}$. Since the area of triangle $F D H$ is $90 \mathrm{~cm}^{2}$, we get $y=45$.
Finally, by the similarity of triangles $X A D$ and $A B C$,
$$
\text { area of } \triangle X B C=\left(\frac{B C}{A D}\right)^{2} y=9 y \text {. }
$$

Hence the area of trapezium $A B C D=8 y=360 \mathrm{~cm}^{2}$.