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Ex 9.3, 16 - In figure, ar(DRC) = ar(DPC) and ar(BDP) - Triangles with same base & same parallel lines

 

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
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Ex 9.3, 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, ΔDPC and ΔDRC 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, ΔADC and ΔBDC 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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