Question

In a trapezium ABCD, side AB || side DC. Diagonals AC and BD cut each other at 0. If AB = 25, DC = 15, OA = 9 then;
i) OC = ?
ii) A(∆COD) : A(∆AOB) = ?
iii) A(∆COD) : A(∆COB) = ?

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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

solution