Question

In the trapezium ABCD AB//CD, AB = 2CD and ar(AOB)=84CM²
find area of triangle COD ​

Answer 1

It is given that ABCD is a trapezium and AB∣∣DC.

In △AOB and △COD,

∠ABO=∠CDO

∠BAO=∠DCO (Alternate angles)

∠AOB=∠COD (Vertically opposite angles)

Therefore, △ABC∼△DEF.

We know that the arc of similar triangles are proportional to squares of their corresponding altitude, therefore with Ar(△AOB)=84 cm

2

and AB=2CD, we have,

Ar(△COD)

Ar(△AOB)

=

CD

2

AB

2

⇒

Ar(△COD)

84

=

CD

2

4CD

2

⇒

Ar(△COD)

84

=4

⇒Ar(△COD)=

4

84

⇒Ar(△COD)=22

Hence, area of △COD is 22 cm square

Answer 2

Given :-

• Diagonals of a trapezium ABCD with AB || CD, intersect each other at point 'O' .
• AB = 2CD .
• ∆AOB area = 84 cm².

To Find :-

• Area of ∆COD = ?

Solution :-

from image we have,

→ ABCD is a trapezium .

→ AB ∣∣ DC .

now, in ∆AOB and ∆COD , we have,

→ ∠ABO = ∠CDO (Alternate angles.)

→ ∠BAO = ∠DCO (Alternate angles.)

→ ∠AOB = ∠COD (Vertically opposite angles.)

so,

→ ∆AOB ~ ∆COD (By AAA Similarity.)

now, we know that,

• If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

then,

→ Ar.∆AOB : Ar.∆COD = AB² : CD²

→ ∆AOB / ∆COD = (AB/CD)²

given that, AB = 2CD

→ 84 / Ar.∆COD = (2CD/CD)²

→ 84 / Ar.∆COD = (2/1)²

→ 84 / Ar.∆COD = 4 / 1

→ Ar.∆COD = 21 cm². (Ans.)

Hence, Area of ∆COD will be 21 cm².

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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