ABCD is a trapezium in which $ A B \| C D $. The diagonals $ A C $ and $ B D $ intersect at $ O . $ If $ O A=6 \mathrm{~cm}, O C=8 \mathrm{~cm} $ find $ \frac{\text { Area }(\Delta A O B)}{\text { Area }(\Delta C O D)} $.

Given:

ABCD is a trapezium in which \( A B \| C D \). The diagonals \( A C \) and \( B D \) intersect at \( O . \)

\( O A=6 \mathrm{~cm}, O C=8 \mathrm{~cm} \).

To do:

We have to find \( \frac{\text { Area }(\Delta A O B)}{\text { Area }(\Delta C O D)} \).

Solution:

$AB \parallel CD$

In $\triangle AOB$ and $\triangle COD$,

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

$\angle BAO=\angle DCO$    (Alternate angles)

Therefore,

$\triangle AOB \sim\ \triangle COD$   (By AA similarity)

We know that,

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

This implies,

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{OA^2}{OC^2}$

$=\frac{(6)^2}{(8)^2}$

$=\frac{36}{64}$

$=\frac{9}{16}$

Therefore,

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{9}{16}$.