Parallelograms

Exercise 9.3 Part 4

Question 13: ABCD is a trapezium with AB||DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar(ADX) = ar(ACY). [Hint: Joint CX]

Answer: ABCD is a trapezium in which AB||DC. The line XY||AC

parallelograms exercise 9.3 answer 13 figure

Here; ar(ACY) = ar(ACX) (Triangles on the same base and between same parallels)
ar(ADX) = ar(ACX)(Triangles on the same base and between same parallels)
Hence, ar(ADX) = ar(ACY) proved.


Question 14: In the given figure, AP||BQ||CR. Prove that ar(AQC) = ar(PBR).

parallelograms exercise 9.3 question 14 figure

Answer: ar(ABQ) = ar(PBQ) (Triangles on the same base and between same parallels)
ar(BQC) = ar(RBQ) (Triangles on the same base and between same parallels)
hence, ar(ABQ) + ar(BQC) = ar(PBQ) + ar(RBQ)
Or, ar(AQC) = ar(PBR) Proved


Question 15: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD) = ar(BOC). Prove that ABCD is a trapezium.

Answer: ar(AOD) = ar(BOC) (Given)

parallelograms exercise 9.3 answer 15 figure

ar(AOD) + ar(DOC) = ar(BOC) + ar(DOC)
Or, ar(ADC) = ar(BDC)
Hence, AB||DC
So, ABCD is a trapezium

Question 16: In the given figure, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

parallelograms exercise 9.3 question 16 figure

Answer: ar(DRC) = ar(DPC) (Given)
Hence, DC||RP
So, DCPR is a trapezium
Now, ar(BDP) = ar(ARC) (Given)
Or, ar(BDP) – ar(DPC) = ar(ARC) – ar(DRC)
Or, ar(ADC) = ar(BDC)
Hence, AB||DC
So, ABCD is a trapezium



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