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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 $$ABCD$$.
1510074_d3a5bb70824b4383be85fc29207c7f9f.png


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 $$120 cm^{2}$$.
1700084_1510074_ans_1a91c67ec7684bb1bc1663df67f9e943.PNG

Mathematics

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