Question

In a trapezium ABCD, AB || CD and AB = 2×CD. If the diagonals AC and BD meet at 0, then area of △AOB is
(A) area (△COD)
(B) 2 area (△COD)
(C) 4 area (△COD)
(D) 3 area (△COD)​

Answer 1

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

solution

Answer 2

Answer:

option a area(∆COD)

Step-by-step explanation:

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