Question

2. Diagonals of a trapezium ABCD with AB II DC intersect each other at the point 0.
If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.




Ans me ​

Answer 1

Given that,

AB = 2 CD

The diagonals of a trapezium ABCD.

AB║DC intersect each other at point O.

We know that,

In the triangles AOB and COD

We know that,

Vertically opposite angles are equal

$\angle DOC = \angle BOA$

Alternate interior angles

$\angle CDO=\angle ABO$

$\angle DCO=\angle BAO$

So, $\triangle AOB\approx \triangle COD$

We need to calculate the ratio of the ares of triangle AOB and COD.

By the similarity rule,

The ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.

Area Δ AOB : Area Δ COD = AB²:CD²

Put the value of AB

Area Δ AOB : Area Δ COD = AB²:CD²

Area Δ AOB : Area Δ COD = (2CD)²:CD²

Area Δ AOB : Area Δ COD = 4CD²:CD²

Area Δ AOB : Area Δ COD = 4:1

Hence, The ratio of the area of triangle AOB and COD is 4:1.

Answer 2

$\huge \mathcal {\fcolorbox{red}{gray}{\green{A}\pink{N}\orange{S}\purple{W}\red{E}\blue{R}}}$

Given that,

AB = 2 CD

The diagonals of a trapezium ABCD.

AB║DC intersect each other at point O.

We know that,

In the triangles AOB and COD

We know that,

Vertically opposite angles are equal

$\angle DOC = \angle BOA$

Alternate interior angles

$\angle CDO=\angle ABO$

$\angle DCO=\angle BAO$

So, $\triangle AOB\approx \triangle COD$

We need to calculate the ratio of the ares of triangle AOB and COD.

By the similarity rule,

The ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.

Area Δ AOB : Area Δ COD = AB²:CD²

Put the value of AB

Area Δ AOB : Area Δ COD = AB²:CD²

Area Δ AOB : Area Δ COD = (2CD)²:CD²

Area Δ AOB : Area Δ COD = 4CD²:CD²

Area Δ AOB : Area Δ COD = 4:1

Hence, The ratio of the area of triangle AOB and COD is 4:1.

HOPE SO IT WILL HELP.....