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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Question : Diagonals of a trapezium $ABCD$ with $AB \parallel CD$ intersect each other at the point $O$. If $AB = 2CD$, then the ratio of the areas of $\triangle AOB$ and $\triangle COD$ is:
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