Question

In the trapezium ABCD, side AB $\parallel$ side DC. Diagonals AC and DB intersect each other at O.

If AB = 15, DC = 10. Find $\frac{OC}{OA}$.

• A. $\frac{OC}{OA}=\frac{1}{3}$
• B. $\frac{OC}{OA}=\frac{7}{3}$
• C. $\frac{OC}{OA}=\frac{2}{3}$
• D. $\frac{OC}{OA}=\frac{5}{3}$

Answer 1

The correct option is C $\frac{OC}{OA}=\frac{2}{3}$

In $△OAB$ and $△OCD$,

$\angle AOB=\angle COD$ (Vertically opposite angles)

$\angle OAB=\angle OCD$ (Alternate angles)

Thus, $△OAB\sim △OCD$ (AA rule)

Hence,

$\frac{OC}{OA}=\frac{CD}{AB}$ (C. S. S.T)

$\frac{OC}{OA}=\frac{10}{15}$

$\frac{OC}{OA}=\frac{2}{3}$