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

Paralleligram ABCD is equal to area of parallelogram ABEF also 2nd figure rectangle ABCD is equal to area of parallelogram ABEF

Rectangles and Parralellograms

Paralleligram 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 Equation

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.

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

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

  • Therefore, Equation (Parallelogram ABCD)……………..equation (2)

  • From (1) and (2), we have Equation

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

  1. Equation

  2. Equation

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, Equation (By construction)……….equation (1)

Thus,

  • Equation and

  • Equation ……….equation (2)

From equations (1) and (2),

  • Equation

  • Equation is a parallelogram.

Now,

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

Also,

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

Equation …………equation (4)

Adding equations (3) and (4),

  • Equation

  • Equation ……..equation (5) Equation

Solution (ii)

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

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

Thus, Equation ……….equation (7)

From equations (6) and (7),

  • Equation

  • 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, Equation ………equation (8)

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

Adding equations (8) and (9),

  • Equation

  • Equation …….equation (10)

From equation (5) and (10) Equation

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