NCERT Class 9 Solutions: Areas of Parallelograms and Triangles (Chapter 9) Exercise 9.2 – Part 2

Give the paralleligrams ABCD is equal to area of parallelogram ABEF also 2nd figure rectangle ABCD is equal to area of parallelogram ABEF

Give the Paralleligrams ABCD and ABEF also2nd Figure Rectangle ABCD

Give the paralleligrams ABCD is equal to area of parallelogram ABEF also 2nd figure rectangle ABCD is equal to area of parallelogram ABEF

Q-3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(APB)=ar(BQC)

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D 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 j Segment j: Segment [A, P] Segment k Segment k: Segment [B, P] Segment l Segment l: Segment [B, Q] Segment m Segment m: Segment [Q, C] Point A A = (-1.78, 4.34) Point A A = (-1.78, 4.34) Point A A = (-1.78, 4.34) Point B B = (2.56, 4.34) Point B B = (2.56, 4.34) Point B B = (2.56, 4.34) Point C C = (2.56, 0.86) Point C C = (2.56, 0.86) Point C C = (2.56, 0.86) Point D D = (-1.72, 0.86) Point D D = (-1.72, 0.86) Point D D = (-1.72, 0.86) Point P P = (1.48, 0.84) Point P P = (1.48, 0.84) Point P P = (1.48, 0.84) Point Q Point Q: Point on d Point Q Point Q: Point on d Point Q Point Q: Point on d

Parallelogram ABCD

Parallelogram ABCD, P and Q are any two point it is laying on the sides DC and AD respectively

Given,

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

APB and parallelogram ABCD are on the same base AB and between the same parallel lines AB and DC. Therefore, ar(APB)=12ar (Parallelogram ABCD)……………equation (1)

  • Similarly, BQC and parallelogram ABCD are on the same base BC and these are between the same parallel lines AD and BC

  • Therefore, ar(BQC)=12ar (Parallelogram ABCD)……………..equation (2)

  • From (1) and (2), we have ar(APB)=ar(BQC)

Q-4 In the figure is a point in the interior of a parallelogram ABCD. Show that

  1. ar(APB)+ar(PCD)=12ar(ABCD)

  2. ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D 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 f Segment f: Segment [A, P] Segment g Segment g: Segment [P, B] Segment h Segment h: Segment [P, C] Segment i Segment i: Segment [P, D] Point A A = (-1.04, 3.62) Point A A = (-1.04, 3.62) Point A A = (-1.04, 3.62) Point B B = (3.32, 3.64) Point B B = (3.32, 3.64) Point B B = (3.32, 3.64) Point C C = (2.04, 1.32) Point C C = (2.04, 1.32) Point C C = (2.04, 1.32) Point D D = (-2.28, 1.32) Point D D = (-2.28, 1.32) Point D D = (-2.28, 1.32) Point P P = (-0.06, 2.68) Point P P = (-0.06, 2.68) Point P P = (-0.06, 2.68)

Parallelogram ABCD

Parallelogram ABCD it’s interior point of P

Solution (i)

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D 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 f Segment f: Segment [A, P] Segment g Segment g: Segment [B, P] Segment h Segment h: Segment [P, C] Segment i Segment i: Segment [P, D] Segment j Segment j: Segment [G, H] Segment P_1 Segment P_1: Segment [E, F] Point A A = (1.72, 1.26) Point A A = (1.72, 1.26) Point A A = (1.72, 1.26) Point B B = (6.44, 1.28) Point B B = (6.44, 1.28) Point B B = (6.44, 1.28) Point C C = (5.1, -1.3) Point C C = (5.1, -1.3) Point C C = (5.1, -1.3) Point D D = (0.44, -1.28) Point D D = (0.44, -1.28) Point D D = (0.44, -1.28) Point P P = (2.72, 0.5) Point P P = (2.72, 0.5) Point P P = (2.72, 0.5) Point G Point G: Point on d Point G Point G: Point on d Point G Point G: Point on d Point H Point H: Point on b Point H Point H: Point on b Point H Point H: Point on b Point E Point E: Point on a Point E Point E: Point on a Point E Point E: Point on a Point F Point F: Point on c Point F Point F: Point on c Point F Point F: Point on c

Parallelogram ABCD With Line GH

Parallelogram ABCD it’s interior point P also line GH is drawn parallel AB passing through P

A line GH is drawn parallel to AB passing through P.In a parallelogram, ABGH (By construction)……….equation (1)

Thus,

  • ADBC and

  • AGBH ……….equation (2)

From equations (1) and (2),

  • ABGH,AGBH

  • ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH. Therefore, ar(ΔAPB)=12ar(ABHG) ………….equation (3)

Also,

In ΔPCD and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.

ar(ΔPCD)=12ar(CDGH) …………equation (4)

Adding equations (3) and (4),

  • ar(ΔAPB)+ar(ΔPCD)=12{ar(ABHG)+ar(CDGH)}

  • ar(APB)+ar(PCD)=12ar(ABCD) ……..equation (5) (ABHG+CDGH=ABCD)

Solution (ii)

A line EF is drawn parallel to AD passing through P.

In a parallelogram, ADEF (by construction)…………equation (6)

Thus, ABCDAEDF ……….equation (7)

From equations (6) and (7),

  • ADEF,ABCD

  • AEFD is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF, ar(ΔAPD)=12ar(AEFD) ………equation (8)

Also, in ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF ar(ΔPBC)=12ar(BCFE) ……..equation (9)

Adding equations (8) and (9),

  • ar(ΔAPD)+ar(ΔPBC)=12[ar(AEFD)+ar(BCFE)]

  • ar(ΔAPD)+ar(ΔPBC)=12[ar(ABCD) …….equation (10)

From equation (5) and (10) ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

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