In the figure given below, $A B C D$ is a trapezium in which $A B ∥ D C$ and $A B = 2 C D$ . Determine the ratio of the areas of $△ A O B$ and $△ C O D$ .

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Solution

In triangle AOB and COD, we have

$∠ A O B = ∠ C O D$ [Vertically opposite angles]
$∠ O B A = ∠ O D C$
and, $∠ O A B = ∠ O C D$ [Alternate angles]

So, by $A A A$ -criterion of similarity, we have $△ A O B \sim △ C O D$

Using the proportionality theorem for similar triangles:
$\frac{A r e a (△ A O B)}{A r e a (△ C O D)} = \frac{{A B}^{2}}{{D C}^{2}}$

$\frac{A r e a (△ A O B)}{A r e a (△ C O D)} = \frac{{(2 D C)}^{2}}{{(D C)}^{2}} = \frac{4}{1}$

Hence, $A r e a (△ A O B) : A r e a (△ C O D) = 4 : 1$ .

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