Class 9 Maths

Parallelograms

Exercise 9.3 Part 3

Question 9: The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see given figure). Show that ar(ABCD) = ar(PBQR).

Hexagon

(Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).

Hexagon

Answer:In triangles ACQ and APQ;
Both the triangles are on the same base, i.e. AQ and between same parallels, i.e. AQ||CP
Hence, ar(ACQ) = ar(APQ)
Now, ar(ACQ) – ar(ABQ) = ar(APQ) – ar(ABQ)
Or, ar(ABC) = ar(PBQ)
As AC is the diagonal of parallelogram ABCD
So, ar(ABC) = ½ ar(ABCD)
As QP is the diagonal of parallelogram BPRQ
So, ar(PBQ) = ½ ar(BPRQ)
Hence, ar(ABCD) = ar(BPRQ) proved

Question 10: Diagonals AC and BD of a trapezium ABCD with AB||DC intersect each other at O. Prove that ar(AOD) = ar(BOC).

parallelogram

Answer: Triangles DAC and CBD lie on the same base and between same parallels
Hence, ar(DAC) = ar(CBD)
Now, ar(DAC) – ar(DOC) = ar(CBD) – ar(DOC)
Or, ar(AOD) = ar(BOC) proved

Question 11: In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. show that

pentagon
  1. ar(ACB) = ar(ACF)
  2. ar(AEDF) = ar(ABCDE)

Answer: Triangles ACB and ACF are on the same base, i.e. CF and between same parallels, i.e. AC and BF
Hence, ar(ACB) = ar(ACF)
Now, ar(ACB) + ar(ACDE) = ar(ACF) + ar(ACDE)
Or, ar(ABCDE) = ar(AEDF) proved

Question 12: A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Answer: ABCD is a quadrilateral. Join A to C and draw BE||AC which intersects DC extended to E.
To prove: ar(ADE) = ar(ABCD)

quadrilateral

BE||AC
Hence, AB = CE
ar(ACB) = ar(CAE) (Triangles on same base and between same parallels)
ar(ACB) + ar(ADC) = ar(CAE) + ar(ADC)
Or, ar(ADE) = ar(ABCD) Proved