Sum

In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD and AB = 2 × CD. If the area of ∆AOB = 84 cm^{2} . Find the area of ∆COD

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#### Solution

In ∆AOB and ∆COD, we have

∠OAB = ∠OCD (alt. int. ∠s)

∠OBA = ∠ODC (alt. int. ∠s)

∴ ∆AOB ~ ∆COD [By AA-similarity]

`\Rightarrow \frac{ar\ (\Delta AOB)}{ar\ (\Delta COD)}=(AB^2)/(CD^2)=(2CD)^2/(CD^2)` [∵ AB = 2 × CD]

`=>(4xxCD^2)/(CD^2)=4`

⇒ ar (∆COD) = 1/4 × ar (∆AOB)

`=>(1/4xx84)cm^2=21cm^2`

Hence, the area of ∆COD is 21 cm^{2} .

Concept: Areas of Similar Triangles

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