NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.2 – Part 3

Alternate Interior angle

Alternate Interior Angle

Alternate Interior angle

Q-5 In the figure, if AB CD,APQ=50° and PRD=127°, find x and y.

Triangle poly1 Triangle poly1: Polygon Q, P, R Angle α Angle α: Angle between F, P, Q Angle α Angle α: Angle between F, P, Q Angle α Angle α: Angle between F, P, Q Angle β Angle β: Angle between P, R, H Angle β Angle β: Angle between P, R, H Angle β Angle β: Angle between P, R, H Angle γ Angle γ: Angle between P, Q, R Angle γ Angle γ: Angle between P, Q, R Angle γ Angle γ: Angle between P, Q, R Angle δ Angle δ: Angle between Q, P, R Angle δ Angle δ: Angle between Q, P, R Angle δ Angle δ: Angle between Q, P, R Segment c Segment c: Segment [Q, P] of Triangle poly1 Segment a Segment a: Segment [P, R] of Triangle poly1 Segment b Segment b: Segment [R, Q] of Triangle poly1 Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [C, D] Point Q Q = (-2.5, 0.52) Point Q Q = (-2.5, 0.52) Point Q Q = (-2.5, 0.52) Point P P = (-0.8, 3.25) Point P P = (-0.8, 3.25) Point P P = (-0.8, 3.25) Point R R = (1.46, 0.52) Point R R = (1.46, 0.52) Point R R = (1.46, 0.52) Point A A = (-3.7, 3.28) Point A A = (-3.7, 3.28) Point A A = (-3.7, 3.28) Point B B = (2.28, 3.28) Point B B = (2.28, 3.28) Point B B = (2.28, 3.28) Point C C = (-3.69, 0.53) Point C C = (-3.69, 0.53) Point C C = (-3.69, 0.53) Point D D = (2.71, 0.55) Point D D = (2.71, 0.55) Point D D = (2.71, 0.55) Point E Point E: Point on g Point E Point E: Point on g Point F Point F: Point on f Point F Point F: Point on f Point G Point G: Point on f Point G Point G: Point on f Point H Point H: Point on g Point H Point H: Point on g

the AB Parallel CD Also Angle of X and Y

The AB parallel CD also angle of x and y

Solution:

Given,

AB CD,APQ=50° and PRD=127°

Now,

x=50° (Alternate interior angles)

PRD+RPB=180° (Angles on the same side of transversal) 127°+RPB=180° RPB=53°

Now, y+50°+RPB=180° (AB is a straight line) y+50°+53°=180°

y+103°=180° y=77°

Q-6 In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AD CD .

Quadrilateral poly1 Quadrilateral poly1: Polygon C_2, D, E, F Quadrilateral poly2 Quadrilateral poly2: Polygon G, H, I, J Segment e Segment e: Segment [E, F] of Quadrilateral poly1 Segment i Segment i: Segment [I, J] of Quadrilateral poly2 Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Vector u Vector u: Vector[C, N] Vector u Vector u: Vector[C, N] Point E E = (10.46, 3.04) Point E E = (10.46, 3.04) Point F F = (5.5, 3.04) Point F F = (5.5, 3.04) Point P Point P: Point on e Point P Point P: Point on e Point P Point P: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point I I = (10.76, 0.1) Point I I = (10.76, 0.1) Point J J = (5.6, 0.1) Point J J = (5.6, 0.1) Point R Point R: Point on i Point R Point R: Point on i Point R Point R: Point on i Point S Point S: Point on i Point S Point S: Point on i Point S Point S: Point on i Point B Point B: Point on e Point B Point B: Point on e Point B Point B: Point on e Point C Point C: Point on i Point C Point C: Point on i Point C Point C: Point on i D text1 = "D" A text3 = "A"

Line of PQ and RS It Is a Parallel

Line of PQ and RS it is a parallel the mirror PQ at B,the reflected ray moves along the path BC and strikes the mirror RSat C and again reflects back along CD

Solution:

Draw ray BLPQandCMRS

Quadrilateral poly1 Quadrilateral poly1: Polygon C_2, D, E, F Quadrilateral poly2 Quadrilateral poly2: Polygon G, H, I, J Angle α Angle α: Angle between K, B, L Angle α Angle α: Angle between K, B, L Angle β Angle β: Angle between L, B, C Angle β Angle β: Angle between L, B, C Angle γ Angle γ: Angle between B, C, M Angle γ Angle γ: Angle between B, C, M Angle δ Angle δ: Angle between M, C, N Angle δ Angle δ: Angle between M, C, N Segment e Segment e: Segment [E, F] of Quadrilateral poly1 Segment i Segment i: Segment [I, J] of Quadrilateral poly2 Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment k Segment k: Segment [K, B] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Segment l Segment l: Segment [B, C] Vector u Vector u: Vector[C, N] Vector u Vector u: Vector[C, N] Vector v Vector v: Vector[B, L] Vector v Vector v: Vector[B, L] Vector w Vector w: Vector[C, M] Vector w Vector w: Vector[C, M] Point E E = (10.46, 3.04) Point E E = (10.46, 3.04) Point F F = (5.5, 3.04) Point F F = (5.5, 3.04) Point P Point P: Point on e Point P Point P: Point on e Point P Point P: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point Q Point Q: Point on e Point I I = (10.76, 0.1) Point I I = (10.76, 0.1) Point J J = (5.6, 0.1) Point J J = (5.6, 0.1) Point R Point R: Point on i Point R Point R: Point on i Point R Point R: Point on i Point S Point S: Point on i Point S Point S: Point on i Point S Point S: Point on i Point B Point B: Point on e Point B Point B: Point on e Point B Point B: Point on e Point C Point C: Point on i Point C Point C: Point on i Point C Point C: Point on i D text1 = "D" A text3 = "A" L text2 = "L" M text4 = "M"

Two Line PQ and RS at a Parallel

Two line PQ and RS at a parallel

BLPQandCMRS and PQ RS

BLCM

LBC=MCB ………..equation(1) (Alternate interiror Angle)

ABL=LBC ………….equation(2) (Angle of incidence = Angle of reflection)

MCB=MCD …………..equation (3) (Angle of incidence = Angle of reflection)

From (1),(2) and (3),we get ABL=MCD ………..equation(4)

Adding (1)and(4),we get

LBC+ABL=MCB+MCD

so,ABC=BCD (ABC=LBC+ABLandBCD=MCB+MCD)

But these are alternate interior angle and they are equal. So, AD CD

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