Ex 9.3, 16 - In figure, ar(DRC) = ar(DPC) and ar(BDP) - Ex 9.3

Ex 9.3, 16 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

 

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Question 16 In figure, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. Given: ar(DRC) = ar(DPC) & ar(BDP) = ar(ARC) To prove: ABCD & DCPR are trapeziums Proof: Given ar(DRC) = ar(DPC) So, DPC and DRC lie on the same base DC and are equal in area & They lie between DC & PR Hence DC PR In DCPR, one pair of opposite sides of quadrilateral DCPR are parallel Hence, DCPR is a trapezium . Now, given that ar(BDP) = ar(ARC) & ar(DPC) = ar(DRC) Subtracting(1) & (2),i.e., (1) (2) ar(BDP) ar(DPC) = ar(ARC) ar(DRC) ar(BDC) = ar(ADC) Now, ADC and BDC lie on the same base DC and are equal in area & They lie between lines DC & AB DC AB Since one pair of opposite sides of quadrilateral ABCD are parallel Hence, ABCD is a trapezium .

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo