Math, asked by sravanthisravvu7, 6 months ago

In the trapezium ABCD, AB//CD and AB = 2CD and diagonals are intersect at 'O'. I
the area of SAOB = 84 cmthen a ACOD =

cm?​

Answers

Answered by Anonymous
1

Answer:

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

2

.

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