Question

# In the adjoining figure, $$ABCD$$ is a parallelogram. $$P$$ is a point on $$BC$$ such that $$BP : PC = 1 : 2$$ and $$DP$$ produced meets $$AB$$ produced at $$Q$$. If area of $$\triangle CPQ = 20 cm^{2}$$, find(i) area of $$\triangle BPQ$$.(ii) area $$\triangle CDP$$.(iii) area of parallelogram $$ABCD$$.

Solution

## From the question it is given that, $$ABCD$$ is a parallelogram.$$BP: PC = 1: 2$$area of $$\triangle CPQ = 20 cm^{2}$$Construction: draw $$QN$$ perpendicular $$CB$$ and Join $$BN$$.Then, area of $$\triangle BPQ$$/area of $$\triangle CPQ = ((\dfrac {1}{2}BP) \times QN)/((\dfrac {1}{2}PC) \times QN)$$$$= BP/PC = \dfrac {1}{2}$$(i) So, area $$\triangle BPQ = \dfrac {1}{2}$$ area of $$\triangle CPQ$$$$= \dfrac {1}{2} \times 20$$Therefore, area of $$\triangle BPQ = 10 cm^{2}$$(ii) Now we have to find area of $$\triangle CDP$$,Consider the $$\triangle CDP$$ and $$\triangle BQP$$,Then, $$\angle CPD = \angle QPD$$ … [because vertically opposite angles are equal]$$\angle PDC = \angle PQB$$ … [because alternate angles are equal]Therefore, $$\triangle CDP \sim \triangle BQP$$ … [$$AA$$ axiom]area of $$\triangle CDP$$/area of $$\triangle BQP = PC^{2}/BP^{2}$$area of $$\triangle CDP$$/area of $$\triangle BQP = 2^{2}/1^{2}$$area of $$\triangle CDP$$/area of $$\triangle BQP = 4/1$$area of $$\triangle CDP = 4 \times area\ \triangle BQP$$Therefore, area of $$\triangle CDP = 4 \times 10$$$$= 40 cm^{2}$$(iii) We have to find the area of parallelogram $$ABCD$$,Area of parallelogram $$ABCD = 2$$ area of $$\triangle DCQ$$$$= 2\ area (\triangle DCP + \triangle CPQ)$$$$= 2 (40 + 20) cm^{2}$$$$= 2 \times 60 cm^{2}$$$$= 120 cm^{2}$$Therefore, the area of parallelogram $$ABCD$$ is $$120 cm^{2}$$.Mathematics

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