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Trapezium

In the follow...

Question

In the following figure, $s e g A B ∥ s e g C D$ . Diagonals $A C$ and $B D$ intersect at point $O$ . If $A O : O C = 1 : 3$ , then $\frac{A (△ A O B)}{A (△ A B D)} =$ ?

\frac{1}{4}

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\frac{1}{9}

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16

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116

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Solution

The correct option is C $\frac{1}{4}$
If $A O : A C : 1 : 3$
$\frac{A r e a o f △ A O B}{A r e a o f △ B O C} = \frac{1}{3}$
Let Area of $△ A O B = x$
$∴$ Area of $△ B O C = 3 x$
Area of $△ (A B C) = 4 x$
Area $△ A B C = △ A B D = 4 x$
$\frac{A r e a o f △ A O B}{A r e a o f △ A B D} = \frac{x}{4 x} = \frac{1}{4}$ .
Hence, option A is correct.

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