Question

ABCD is a trapezium in which AB || DC and AB = 2DC. If the
diagonals of the trapezium intersect each other at a point o, find the
ratio of the areas of triangle AOB and triangleCOD.​

Answer 1

Given that,

In trapezium ABCD, AB || DC and AB = 2DC.

To find,

The ratio of the areas of ∆ AOB and ∆ COD.

Solution,

In ∆ AOB and ∆ COD,

∠AOB = ∠COD [ vertically opposite angles]

∠OAB = ∠OCD [ Alternative interior angles]

Therefore, by AA similarity,

∆AOB ∽ ∆COD.

We know that,

the ratio of the areas of two similar triangles is equal to the ratio of the square of the corresponding sides.

$\sf\dfrac{ar(\triangle \: AOB)}{ar(\triangle \: COD)} = \dfrac{AB^2}{DC^2}$

we know that, AB = 2DC, so,

$\sf\dfrac{ar(\triangle \: AOB)}{ar(\triangle \: COD)} = \dfrac{(2 \times DC)^2}{DC^2}$

$\sf\dfrac{ar(\triangle \: AOB)}{ar(\triangle \: COD)} = \dfrac{4 \times DC^2}{DC^2}$

$\sf\dfrac{ar(\triangle \: AOB)}{ar(\triangle \: COD)} = \dfrac{4}{1}$

Thus, ar(∆AOB) : ar(∆COD) = 4:1