Question : In trapezium $ABCD$, $AB \parallel CD$ and $AB = 2CD$. Its diagonals intersect at $O$. If the area of $\triangle AOB=84\;\mathrm{cm^2}$ then the area of $\triangle COD$ is equal to:
Option 1: 72 cm2
Option 2: 21 cm2
Option 3: 42 cm2
Option 4: 26 cm2
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Correct Answer: 21 cm 2
Solution :
In trapezium $ABCD$, $AB \parallel CD, AB = 2CD$ and area of $\Delta AOB=84\;\mathrm{cm^2}$.
In $\triangle AOB$ and $\triangle COD$,
$\angle AOB=\angle COD$ (Vertically opposite angles)
$\angle ABO=\angle CDO$ (Alternate angles)
$\angle BAO=\angle DCO$ (Alternate angles)
So, $\triangle AOB\sim\triangle COD$
Using the theorem, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\frac{\operatorname{Area of }\Delta AOB}{\operatorname{Area of }\Delta COD}=\frac{AB^2}{CD^2}$
⇒ $\frac{84}{\operatorname{Area of }\Delta COD}=\frac{4CD^2}{CD^2}=4$
⇒ $\operatorname{Area of }\triangle COD=21$ cm
2
Hence, the correct answer is 21 cm
2
.
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