In trapezium ABCD,AB || CD and AB = 2 , CD. I | vedclass

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(A) In trapezium $ABCD$,since $AB || CD$,$\Delta AOB$ is similar to $\Delta COD$ by $AA$ similarity criterion ($\angle OAB = \angle OCD$ and $\angle O

Vedclass · Accepted Answer

$21$

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(A) In trapezium $ABCD$,since $AB || CD$,$\Delta AOB$ is similar to $\Delta COD$ by $AA$ similarity criterion ($\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$ as alternate interior angles).The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.$\frac{	ext{Area}(\Delta AOB)}{	ext{Area}(\Delta COD)} = \left( \frac{AB}{CD} ight)^2$Given $AB = 2 \, CD$,so $\frac{AB}{CD} = 2$.$\frac{84}{	ext{Area}(\Delta COD)} = (2)^2 = 4$$	ext{Area}(\Delta COD) = \frac{84}{4} = 21 \, cm^2$.

In trapezium $ABCD$,$AB || CD$ and $AB = 2 \, CD$. Its diagonals intersect at $O$. If the area of $\Delta AOB = 84 \, cm^2$,then the area of $\Delta COD$ is equal to (in $cm^2$)

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