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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 D)}{\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 \( \frac{\text { Area }(\Delta A O D)}{\text { Area }(\Delta C O D)} \).
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,
Area of a triangle of base $b$ and height $h$ is $\frac{1}{2}\times b \times h$.
This implies,
$\frac{ar(\triangle AOD)}{ar(\triangle COD)}=\frac{\frac{1}{2}\times AO \times DP}{\frac{1}{2}\times CO \times DP}$
$=\frac{AO}{CO}$
$=\frac{6}{8}$
$=\frac{3}{4}$
Therefore,
$\frac{ar(\triangle AOD)}{ar(\triangle COD)}=\frac{3}{4}$.
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