Question

find and write 3 pairs of triangle having same area from the figure given AB parallel to DC. The diagonals AC and BD intersect at P
​

Answer 1

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

Hope it helps you