7. $A B C D$ is a trapezium inscribed in a circle centred at $O$. It is given that $A B \| C D, \angle C O D=$ $3 \angle A O B$, and $\frac{A B}{C D}=\frac{2}{5}$.

Find the ratio
$$
\frac{\text { area of } \triangle B O C}{\text { area of } \triangle A O B} \text {. }
$$
(A) $\frac{3}{2}$;
(B) $\frac{7}{4}$;
(C) $\frac{\sqrt{3}}{\sqrt{2}}$;
(D) $\frac{\sqrt{5}}{2}$;
(E) $\frac{\sqrt{7}}{\sqrt{2}}$
7. Answer: (A).

Let $\angle A O B=\alpha$ and $\angle A D O=\beta$. Then $\frac{1}{2}(\alpha+3 \alpha)=2 \beta$; that is, $\alpha=\beta$. Given that
$$
\frac{5}{2}=\frac{\sin (3 \alpha / 2)}{\sin (\alpha / 2)}=\frac{3 \sin (\alpha / 2)-4 \sin ^{3}(\alpha / 2)}{\sin (\alpha / 2)}=3-4 \sin ^{2}(\alpha / 2) .
$$

Then $\sin ^{2}(\alpha / 2)=\frac{1}{8}$. Hence,
$$
\frac{S_{\triangle B O C}}{S_{\triangle A O B}}=\frac{\sin (\pi-2 \alpha)}{\sin \alpha}=2 \cos \alpha=2\left(1-2 \sin ^{2}(\alpha / 2)\right)=\frac{3}{2} .
$$