# 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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Updated on: 10-Oct-2022

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