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Trapeziums and kites - class-V
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Diagonals of trapezium $ABCD$ with $AB\parallel DC$ intersect each other at the point $O$. If $AB=2CD$, find the ratio of the areas of triangles $AOB$ and $COD$.
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A
$4:1$
💡 Explanation:
Given: $AB \parallel CD$ and $AB = 2 CD$
In $\triangle OAB$ and $\triangle OCD$
$\angle AOB = \angle COD$ (Vertically opposite angles)
$\angle ODC = \angle OBA$ (Alternate angles)
$\angle OCD = \angle OAB$ (Alternate angles)
thus, $\triangle OAB \cong \triangle OCD$ (AAA rule)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{AB^2}{CD^2}$ (Similar triangle Property)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{(2 CD)^2}{CD^2}$
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = 4 : 1$