NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.1 – Part 2

Q-3 In the figure, PQR=PRQ , then prove that PQS=PRT.

Triangle poly1 Triangle poly1: Polygon P, Q, R Angle α Angle α: Angle between P, Q, R Angle α Angle α: Angle between P, Q, R Angle β Angle β: Angle between P, R, Q Angle β Angle β: Angle between P, R, Q Segment f Segment f: Segment [P, Q] of Triangle poly1 Segment d Segment d: Segment [Q, R] of Triangle poly1 Segment e Segment e: Segment [R, P] of Triangle poly1 Vector u Vector u: Vector[A, T_1] Vector u Vector u: Vector[A, T_1] Vector v Vector v: Vector[A, S_1] Vector v Vector v: Vector[A, S_1] Point P P = (0.44, 5.42) Point P P = (0.44, 5.42) Point P P = (0.44, 5.42) Point Q Point Q: Point on v Point Q Point Q: Point on v Point Q Point Q: Point on v Point R Point R: Point on u Point R Point R: Point on u Point R Point R: Point on u Point S Point S: Point on v Point S Point S: Point on v Point S Point S: Point on v Point T Point T: Point on u Point T Point T: Point on u Point T Point T: Point on u

Point of P, Q, R, S, T and Triangle PQR

Prove angle PQS equal angle PRT

Solution:

In figure give point of P, Q, R, S, T and PQR=PRQ

To prove,

PQS=PRT

Equation,

PQR+PQS=180° (Linear Pair)

PQS=180°PQR …………equation (1)

Also,

PRQ+PRT=180° (Linear Pair)

PRT=180°PRQ

PRQ=180°PQR ………..equation (2) (PQR=PRQ)

From (i) and (ii)

PQS=PRT =180°PQR

Therefore, PQS=PRT

Q-4 In the figure, if x+y=w+z then proves that AOB is a line.

Angle α Angle α: Angle between C, O, A Angle α Angle α: Angle between C, O, A Angle α Angle α: Angle between C, O, A Angle β Angle β: Angle between A, O, D Angle β Angle β: Angle between A, O, D Angle β Angle β: Angle between A, O, D Angle γ Angle γ: Angle between D, O, B Angle γ Angle γ: Angle between D, O, B Angle γ Angle γ: Angle between D, O, B Angle δ Angle δ: Angle between B, O, C Angle δ Angle δ: Angle between B, O, C Angle δ Angle δ: Angle between B, O, C Vector u Vector u: Vector[O, B_1] Vector u Vector u: Vector[O, B_1] Vector v Vector v: Vector[O, C_1] Vector v Vector v: Vector[O, C_1] Vector w Vector w: Vector[O, D_1] Vector w Vector w: Vector[O, D_1] Vector a Vector a: Vector[O, E] Vector a Vector a: Vector[O, E] Point O O = (0.16, 2.1) Point O O = (0.16, 2.1) Point O O = (0.16, 2.1) Point C Point C: Point on w Point C Point C: Point on w Point C Point C: Point on w Point B Point B: Point on u Point B Point B: Point on u Point B Point B: Point on u Point D Point D: Point on a Point D Point D: Point on a Point D Point D: Point on a Point A Point A: Point on v Point A Point A: Point on v Point A Point A: Point on v

Point of a, B, C, D and Angle of W, X, Y, Z

Point of A, B, C, D and angle of w, x, y, z. Prove AOB in a line

Given, point of A, B, C, D and angle of w, x, y, z

x+y=w+z

Equation,

x+y+w+z=360° (Angles around a point)

(x+y)+(w+z)=360°

(x+y)+(x+y)=360°(Givenx+y=w+z)

2(x+y)=360°

(x+y)=180°

Hence, x+y makes a linear pair. Therefore, AOB is a staright line.

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