Class 9 Maths Chapter 6 Question Answers - Lines and Angles

Q1:  In the adjoining  figure, AB || CD. If ∠ APQ = 54° and ∠ PRD = 126°, then find x and y.
 Solution: ∵ AB || CD and PQ is a transversal, then interior alternate angles are equal.

⇒ ∠ APQ = ∠ PQR  [alt. interior angles]
⇒ 54° = x       [∵ ∠ APQ = 54° (Given)]
Again, AB || CD and PR is a transversal, then ∠ APR = ∠ PRD    [Interior alternate angles]
But ∠ PRD = 126°        [Given]
∴ ∠ APR = 126°
Now, exterior∠ PRD + ∠ PRQ = 180°
⇒ 126°+∠ PRQ = 180°
⇒∠ PRQ= 180°-126° = 54°

In triangle PQR ,

∠ PQR + ∠ PRQ + ∠ QPR =  180° ( Angle sum property) 

54°+ 54° +∠ QPR= 180°

108+ ∠ QPR =180°

∠ QPR =180°-108° = 72° 
Thus, x = 54° and y = 72°.

Q2: In the adjoining figure AB || CD || EG, find the value of x.
 Solution: Let us draw FEG || AB || CD through E.
Now, since FE || AB and BE are transversals,
∴ ∠ ABE + ∠ BEF = 180°
[Interior opposite angles]

⇒ 127° + ∠ BEF = 180°  [co int. angles] 
⇒ ∠ BEF = 180° - 127° = 53°
Again, EG || CD and CE is a transversal.
∴ ∠ DCE + ∠ CEG = 180°          [Interior opposite angles]
⇒ 108° + ∠ CEG = 180° ⇒ ∠ CEG = 180° - 108° = 72°
Since FEG is a straight line, then
⇒ ∠BEF + ∠BEC + ∠CEG = 180°                 [Sum of angles at a point on the same side of a line = 180°]
⇒ 53° + x + 72° = 180°
⇒ x = 180° - 53° - 72°
= 55°
Thus, the required measure of x = 55°.

Q3: AB and CD are parallel, and EF is a transversal. If ∠BEF = 70°, find ∠EFD and ∠CFE

Ans: Since AB || CD and EF is a transversal, by the Corresponding Angles Theorem:

∠BEF = ∠EFD

Thus, ∠EFD = 70°.

By the Linear Pair Axiom:

∠EFD + ∠CFE = 180°

70° + ∠CFE = 180°

∠CFE = 110°.

So, ∠EFD = 70° and ∠CFE = 110°.

Q4:  If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles enclose a rectangle.

Ans: 

Two parallel lines AB and CD are intersected by a transversal L at P and R

respectively

PQ, RQ, RS and PS are bisectors of ∠APR, ∠PRC, ∠PRD and ∠BPR respectively.

Since AB || CD and L is a transversal

∠APR = ∠PRD ( alt. interior angles)

∠APR = 1/2(∠PRD)  = ∠QPR = ∠PRS

these are alternate interior angles.

QP || RS. Similarly QR || PS.

PQRS is a parallelogram.

Also ray PR stands on AB

∠APR + ∠BPR = 180° ( linear pair)

∠APR + (1/2) ∠BPR = 90°

∠QPR + ∠SPR = 90° 

∠QPS = 90°

Therefore PQRS is a parallelograrn, one of whose angle is 90°.

Hence PORS satisfies all the properties of being a rectangle.

Hence PQRS is a rectangle

Q5: In the given figure, AOC is a line, find x. 

Ans: AOC is a straight line 

∠AOB + ∠BOC= 180°

60 + 3x = 180°

3x = 180 - 60 

x = 120/ 3 

x= 40°