# 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

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).

**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)

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.

**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