Question

Solve pls pls.. fast .

Answer 1

Given:

abcd is a parallelogram, p is a mid point of ab . bd and cp intersect at q such that cq:qp =3:1. ar(triangle pbq ) = 10 cm^2

To find :

the area of parallelogram abcd.

Soln :

CQ / QP = 3/1

or

3QP=CQ ..................................... Eq 1

=> 1/2 * PB * QP = 10 cm^2

=> 1/2 * PB * CQ / 3 = 10 cm^2     { from eq 1}

=> 1/2 * PB * CQ = 3 * 10 cm^2

=> 1/2 * PB * CQ = 30 cm^2

 Area ( triangle PBQ) + Area (triangle CBQ) = Area (triangle PBC)

=> 10 cm^2 + 30 cm^2 = 40 cm^2

therefore Area (triangle PBC) = 40 cm^2

since P is the mid point of AB

therefore CP is the median of traingle ABC

Area ( triangle PBC ) = Area ( triangle APC)

or Area ( triangle APC) = 40 cm^2

thus area of triangle ABC = 40 + 40 = 80 cm^2

area of parallelogram abcd = 160 cm^2

hope this answer was helpful...........

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