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.

Give point of P,Q,R,S,T and ∠PQR = ∠PRQ so prove ∠PQS = ∠PRT.

Give Point of P,Q,R,S,T and ∠PQR = ∠PRQ

Give point of P,Q,R,S,T and ∠PQR = ∠PRQ so prove ∠PQS = ∠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 prove that AOB is a line.

Give point of A,B,C,D and angle of w,x,y,z then x+y=w+z so prove AOB in a line

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

Give point of A,B,C,D and angle of w,x,y,z then x+y=w+z so prove AOB in a line

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

To Prove,AOB is a line orx+y=180° (linear pair.)

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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