Question

= B 12. In the given figure, ABCD is a trapezium in which AB || DC and diagonals AC and BD intersect at point 0. If BO: OD = 4:7; find : A (i) ArAOD : ArAOB (ii) ArAOB : ArACB (iii) ArDOC: ArAOB (iv) ArABD : ArBOC ​

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)=OC2OA2                [ By area theorem ]

⇒  ar(△COD)ar(△AOB)=(8)2(6)2

⇒  ar(△COD)ar(△AOB)=6436=169

(b)   Draw DP⊥AC

ar(△COD)ar(△AOD)=21×CO×DP21×