Question

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

Answer 1

Answer:

Given,

ABCD is a trapezium with AB∣∣CD .......(1)

And

AB=2CD ......(2)

In the triangles AOB and COD,

∠DOC=∠BOA [vertically opposite angles are equal]

∠CDO=∠ABO [alternate interior angles ]

∠DCO=∠BAO

Thus,

△AOB≈△COD

By the similarity rule, the ratio of the areas of the similar triangles is the ratio of the square of corresponding sides.

therefore

Area (△AOB):Area (△COD)=AB

2

:CD

2

Area (△AOB):Area (△COD)=(2CD)

2

:CD

2

Area (△AOB):Area (△COD)=4CD

2

:CD

2

Area (△AOB):Area (△COD)=4:1

Hence, this is the answer.