In a trapezium ABCD, O is the point of intersection of AC and BD, AB ∥ CD and AB = 2 × CD. If area of AOB = 84 cm², find the area of ΔCOD.

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Solution

In Δ AOB and Δ COD, we have:
$∠$ OAB = $∠$ OCD [ Alternate interior angles]
$∠$ OBA = $∠$ ODC [ Alternate interior angles]
∴ ΔAOB ∼ Δ COD [ By AA similarity]
$\frac{a r △ A O B}{a r △ C O D} = \frac{A B^{2}}{C D^{2}} = \frac{{(2 C D)}^{2}}{C D^{2}}$ [ Since AB = 2 × CD]
$= \frac{4 \times C D^{2}}{C D^{2}} = 4$
$\Rightarrow a r (△ C O D) = \frac{1}{4} \times a r (△ A O B) = (\frac{1}{4} \times 84) {cm}^{2} = 21 {cm}^{2}$
Hence, area of ΔCOD is 21 cm².

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In a trapezium ABCD, O is the point of intersection of AC and BD, AB ∥ CD and AB = 2 × CD. If area of AOB = 84 cm2, find the area of ΔCOD.

In a trapezium ABCD, O is the point of intersection of AC and BD, AB ∥ CD and AB = 2 × CD. If area of AOB = 84 cm², find the area of ΔCOD.