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In quadrilateral ABCD, diagonals $AC$ and BD intersect each other at point O such that $A O O C = B O O D = 12$ . Find the value of $a r Δ A O B a r Δ C O D$ .

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1:3

2:3

1:4

1:2

answer is C.

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Detailed Solution

It is given that,

\frac{AO}{OC} = \frac{BO}{OD} = \frac{1}{2}

We know that if the ratios of corresponding angles are equal then the triangles are similar.
Hence, we can write that

ΔOAB \sim ΔOCD

.
Let us draw the figure.
In the given figure, ABCD is a trapezium in which AB║DC and its diagonals intersect at O. If AO = (5x – 7), OC = (2x + 1) - Sarthaks eConnect | Largest Online Education Community

In the given figure, ABCD is a trapezium in which AB║DC and its diagonals intersect at O. If AO = (5x – 7), OC = (2x + 1) - Sarthaks eConnect | Largest Online Education Community

Since

Δ O A B ~ Δ O C D

we can write that,

A O O C = B O O D = A B D C A B D C = 12 4 D C = 12 D C = 8 c m

The ratio of area of similar triangles is equal to the square of the ratio of corresponding sides of similar triangles.
Hence,

a r Δ A O B a r Δ C O D = 14

The value is 1:4.
Hence, the correct option is 3.

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