Question

In the given fig, ar(DRC) and ar(BDP) = (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

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Answer 1

Answer:

Data: in fig. ar.(DRC) = ar.(DPC) and ar.(BDP) = ar.(ARC). To Prove: (i) ABCD is a trapezium (ii) DCPR is a trapezium. Proof: (i) ar.(∆BDP) = ar.(∆ARC) (Data) ar.(∆DPC) + ar.(∆DCB) = ar.(∆DCR) + ar.(∆DCA) But, ∆DRC = ∆DPC. (Data) ∴ ar.(∆DCB) = ar.(∆DCA) These are on same base DC and in between straight lines DC and AB. ∴ DC || AB. Now, one pair of opposite sides of quadrilateral ABCD are parallel, hence ABCD is a trapezium. (ii) ∆DRC and ∆DPC are on the same base DC and in between DC and RP and they are equal in area. ∴ DC || RP. ∴ Hence ‘Quadrilateral’ DCPR is a trapezium.

Step-by-step explanation:

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