NCERT Class 9 Solutions: Areas of Parallelograms and Triangles (Chapter 9) Exercise 9.2 – Part 3

The figure below illustrates that area of triangle is ½ that of a parallelogram, if they are between same parallel lines and have same base.

Area of triangle is 1/2 that of a parallelogram, if they are beween same parallel lines and have same base.

Area of Triangle = 1/2bh and Parallelogram Is Bh

Area of triangle is 1/2 that of a parallelogram, if they are beween same parallel lines and have same base.

Q-5 In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that

  1. ar(PQRS)=ar(ABRS)

  2. ar(AXS)=12ar(PQRS)

Quadrilateral poly1 Quadrilateral poly1: Polygon P, B, R, S Triangle poly2 Triangle poly2: Polygon A, X, S Segment a Segment a: Segment [P, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, R] of Quadrilateral poly1 Segment c Segment c: Segment [R, S] of Quadrilateral poly1 Segment d Segment d: Segment [S, P] of Quadrilateral poly1 Segment d_1 Segment d_1: Segment [A, X] of Triangle poly2 Segment e Segment e: Segment [X, S] of Triangle poly2 Segment f Segment f: Segment [S, A] of Triangle poly2 Segment g Segment g: Segment [R, Q] Point P P = (-2.32, 3.26) Point P P = (-2.32, 3.26) Point P P = (-2.32, 3.26) Point B B = (3.52, 3.18) Point B B = (3.52, 3.18) Point B B = (3.52, 3.18) Point R R = (2.14, 0.42) Point R R = (2.14, 0.42) Point R R = (2.14, 0.42) Point S S = (-2.96, 0.32) Point S S = (-2.96, 0.32) Point S S = (-2.96, 0.32) Point A Point A: Point on a Point A Point A: Point on a Point A Point A: Point on a Point X Point X: Point on b Point X Point X: Point on b Point X Point X: Point on b Point Q Point Q: Point on a Point Q Point Q: Point on a Point Q Point Q: Point on a

Parallelogram PQRS and ABRS

Parallelogram PQRS and ABRS are on the same base SR, and SR and PB are the same parallel lines

Solution (i): ar(PQRS)=ar(ABRS)

  • Parallelogram PQRS and ABRS are on the same base SR

  • And, SR and PB are the same parallel lines. Therefore they have the same base length and height. Thus,

    ar(PQRS)=ar(ABRS) …………equation (1)

Solution (ii): ar(AXS)=12ar(PQRS)

  • In triangle AXS and parallelogram ABRS are on the same base AS

  • And, AS and BR are the same parallel lines.

ar(ΔAXS)=12ar(ABRS)

2ar(ΔAXS)=ar(ABRS) ……………..equation (2)

From (1) and (2), ar(PQRS)=2ar(ΔAXS)

ar(ΔAXS)=12ar(PQRS)

Q-6 A farmer was having a field in the form of a parallelogram ABCD. She took any point P on CD and joined it to points A and B. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Triangle poly2 Triangle poly2: Polygon A, P, B Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment b_1 Segment b_1: Segment [A, P] of Triangle poly2 Segment a_1 Segment a_1: Segment [P, B] of Triangle poly2 Segment e Segment e: Segment [B, A] of Triangle poly2 Point A A = (-1.66, 4) Point A A = (-1.66, 4) Point A A = (-1.66, 4) Point B B = (1.88, 4) Point B B = (1.88, 4) Point B B = (1.88, 4) Point C C = (0.98, 1) Point C C = (0.98, 1) Point C C = (0.98, 1) Point D D = (-2.52, 1) Point D D = (-2.52, 1) Point D D = (-2.52, 1) Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c

Parallelogram ABCD

Parallelogram ABCD and point P on CD joined to points A and B

  • The field is divided into three parts.

  • The three parts triangles, namely, ΔADP,ΔABPandΔBAC .

  • Area of ΔADP+ΔAPB+ΔBPC=AreaofABCD …………..equation (1)

  • Since, triangle ΔAPB and parallelogram ABCD on the same base, area of ΔAPB=12areaofABCD

From (1) and (2),

  • Area of ΔADP+ΔAPB+ΔBPC=AreaofABCD

  • Area of ΔADP+12areaofABCD+ΔBPC=AreaofABCD

  • Area of ΔADP+ΔBPC=AreaofABCD12areaofABCD

  • Area of ΔADP+ΔBPC=12areaofABCD ………equation (3)

Now ΔADP+ΔBPC form half the area of ABCD and ΔAPB form the other half, therefore, the farmer should sow wheat in one of these half and pulses in other half.

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