ABCD is a trapezium in which AB || DC and AB = 2DC. If thediagonals of

1.	ABCD is a trapezium in which AB \|\| DC and AB = 2DC. If thediagonals of the trapezium intersect each other at a point o, find theratio of the areas of triangle AOB and triangleCOD.
Answer» 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

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

Answer»

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