In a trapezium ABCD, AB || CD, the diagonals AC and BD intersect at 'P'. If
AB: CD = 2:1, then area of ACPD : area of AAPB =
3) 1.2
Answers
Answer:
ABCD is a trapezium in which AB∥CD.
The diagonals AC and BD intersect at O. OA=6 cm and OC=8cm
In △AOB and △COD,
∠AOB=∠COD [ Vertically opposite angles ]
∠OAB=∠OCD [ Alternate angles ]
∴ △AOB∼△COD [ By AA similarity ]
(a)
ar(△COD)
ar(△AOB)
=
OC
2
OA
2
[ By area theorem ]
⇒
ar(△COD)
ar(△AOB)
=
(8)
2
(6)
2
⇒
ar(△COD)
ar(△AOB)
=
64
36
=
16
9
(b) Draw DP⊥AC
ar(△COD)
ar(△AOD)
=
2
1
×CO×DP
2
1
×AO×DP
[ By area theorem ]
⇒
ar(△COD)
ar(△AOD)
=
CO
AO
⇒
ar(△COD)
ar(△AOD)
=
8
6
=
4
3
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