Question

In a trapezium ABCD, diagonals AC and BD intersect at O. If AB = 3CD, then find ratio of areas of triangles COD and AOB.

Answer 1

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Answer 2

The ratio of area of triangles COD and AOB is 1:9.

Step-by-step explanation:

It is given that,

In a trapezium ABCD,  

Diagonals AC and BD intersect at point O.

$AB = 3 DC$ ….. (i)

Let’s consider ∆AOB and ∆COD, we have

$\angle AOB = \angle COD$ ….. [since AB // CD, therefore angle AOB & angle COD are vertically opposite angles]

$\angle ABO = \angle CDO$ …… [alternate angles, AB//CD & BD is transversal line]

∴ By AA similarity, $\triangle AOB$ ~ $\triangle COD$

We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

∴ [Area of $\triangle COD$] / [Area of $\triangle AOB$] is,

$=[\frac{AC}{AB} ]^2$  

$=[\frac{CD}{3CD} ]^2$…….. [substituting from (i)]

$=[\frac{1}{3}]^2$

$=\frac{1}{9}$

Thus, the ratio of the area of triangles COD and AOB is $1:9$.

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