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Class 9 NCERT Solutions Chapter 6 Maths

Chapter 6: Lines and Angles

TEXTBOOK’S EXERCISE – 6.2

1. In the figure, find the values of x and y and then show that AB || CD.

Sol. In the given figure, a transversal intersects two lines AB and CD such that

x + 50° = 180° [Linear pair axiom]

⇒ x = 180° – 50° = 130°

y = 130° [Vertically opposite angles]

Therefore, ∠x = ∠y = 130° [Alternate angles]

∴ AB || CD [Converse of alternate angles axiom] Proved.

2. In the figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Sol. In the given figure, AB || CD, CD || EF and y : z = 3 : 7.

Let y = 3a and z = 7a

∠DHI = y [Vertically opposite angles]

∠DHI + ∠FIH = 180° [Interior angles on the same side of the transversal]

⇒ y + z = 180°.

⇒ 3a + 7a = 180°

⇒ 10a = 180° ⇒ a = 18°

∴ y = 3 × 18° = 54° and z = 18° × 7 = 126°

Also, x + y = 180°

⇒ x + 54° = 180°

∴ x = 180° – 54° = 126°

Hence, x = 126°.

Sol. In the given figure, AB || CD, EF ⊥ CD and ∠GED = 126°

∠AGE = ∠GED [Alternate angles]

∴ ∠AGE = 126°

Now, ∠GEF = ∠GED – ∠DEF

= 126° – 90° = 36° [Since ∠DEF = 90°]

Also, ∠AGE + ∠FGE = 180° [Linear pair axiom]

⇒ 126° + FGE = 180°

⇒ ∠FGE = 180° – 126° = 54°.

4. In the figure, if PQ || ST, ∠PQR = 110° and ∠ RST = 130°, find ∠QRS.

Sol. Extend PQ to Y and draw LM || ST through R.

∠TSX = ∠QXS [Alternate angles]

⇒ ∠QXS = 130°

∠QXS + ∠RXQ = 180° [Linear pair axiom]

⇒ ∠RXQ = 180° – 130° = 50° …(1)

∠PQR = ∠QRM [Alternate angles]

⇒ ∠QRM = 110° …(2)

∠RXQ = ∠XRM [Alternate angles]

⇒ ∠XRM = 50° [By (1)]

∠QRS = ∠QRM – ∠XRM

= 110° – 50° = 60°

5. In the figure, if PT ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Sol. In the given figure, lines PQ ⊥ PS, PQ || SR,

∠SQR = 28° and ∠QRT = 65°

∠PQR = ∠QRT [Alternate angles]

⇒ x + 28° = 65°

⇒ x = 65° – 28° = 37°

In ΔPQS,

∠SPQ + ∠PQS + ∠QSP = 180° [Angle sum property of a triangle]

⇒ 90° + 37° + y = 180° [Since PQ ⊥ PS, ∠PQS = x = 37° and ∠QSP = y]

⇒ 127° + y = 180°.

⇒ y = 180° – 127° = 53°

Hence, x = 37° and y = 53°

6. In the figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1/2 ∠ QPR.

Sol. Exterior ∠PRS = ∠PQR + ∠QPR [Exterior angle property]

Therefore, 1/2 ∠PRS = 1/2 ∠PQR + 1/2 ∠QPR

⇒ ∠TRS = ∠TQR + 1/2 ∠QPR …(i)

But in ΔQTR,

Exterior ∠TRS = ∠TQR + ∠QTR …(ii) [Exterior angle property]

Therefore, from (i) and (ii)

∠TQR + ∠QTR = ∠TQR + 1/2 ∠QPR

⇒ ∠QTR = 1/2 ∠QPR Proved.
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