Question

In the Fig. ABCD is a parallelogram and AP: PB = 3: 5. Calculate:
(i) ar (APBN): ar (trapezium APND)
(ii) PN: BC and ar (APMN): ar (ABMC)
(iii) BM : DM and BM : BD
iv) ar (A BMC): ar (A DMC) and ar (triangle BDA): ar (ADMC)​

Answer 1

Given : ABCD is a parallelogram and AP: PB = 3: 5.

To Find : Area of ΔPBN  :  ar ( trapezium APND)

PN : BC

Solution:

ABCD is parallelogram

AD || BC

PN || AD || BC  given

∠A = ∠P  ( corresponding angle)

∠D = ∠N ( corresponding angle)

∠B = ∠B

ΔABD ≈ ΔPBN

=> PN / AD  =  PB/AB

AP : PB = 3 : 5

=> AP : AB = 3 : 8   & PB : AB = 5 : 8

=> PN / AD  =  5/8

=> PN = 5AD/8

Area of ΔPBN =  (5/8)²Area of ΔABD

=> Area of ΔPBN = (25/64) Area of ΔABD

Area of ΔABD = (64/25) Area of ΔPBN

ar ( trapezium APND) =  Area of ΔABD  -  Area of ΔPBN

=> ar ( trapezium APND) =  Area of ΔABD  -  (25/64) Area of ΔABD

=>  ar ( trapezium APND) = (39/64) Area of ΔABD

      Area of ΔPBN = (25/64) Area of ΔABD

=> Area of ΔPBN  :  ar ( trapezium APND) = 25 : 39

PN / AD  =  5/8

AD = BC

=> PN : BC = 5 : 8

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