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?
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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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