**NCERT Solutions for Class 9
Maths Chapter 6 Lines and Angles Ex 6.2**

NCERT Solutions for Class
9 Maths Chapter 6 Lines
and Angles Ex 6.2
are the part of NCERT
Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 6 Lines
and Angles Ex 6.2.

**NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.1****NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.2****NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3**

**Ex 6.2 Class 9 Maths Question
1.**

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

**Solution:**

Here, l is a straight line.

So, x + 50° = 180° [By linear pair of angles]

⇒ x = 180° – 50° = 130°

Now, line l and CD intersect each other at a point.

So, y = 130° [Vertically opposite angles]

Thus, x = y = 130° [Alternate interior angles]

Hence, AB || CD [By theorem 2]

**Hence proved.**

**Ex 6.2 Class 9 Maths**** ****Question 2.****
**In figure,
if AB || CD, CD || EF and y : z = 3 : 7, find x.

**Solution:
**We
have, AB || CD and CD || EF [Given]

∴ AB || EF

∴ x = z [Alternate interior angles] ….(1)

Again, AB || CD

⇒ x + y = 180° [Co-interior angles]

⇒ z + y = 180° … (2) [By (1)]

But y : z = 3 : 7

Let y = 3k and z = 7k

7k + 3k = 180° [By (2)]

⇒ 10k = 180°

⇒ k = 18°

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

⇒ x = 126° [From (1)]

**Ex 6.2 Class 9 Maths**** ****Question 3.****
**In figure,
if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

**Solution:****
**We have AB
|| CD and GE is a transversal.

∴ ∠AGE = ∠GED [Alternate interior angles]

But ∠GED = 126° [Given]

∴ ∠AGE = 126°

Also, ∠GEF + ∠FED = ∠GED

or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]

∠GEF = 126° – 90° = 36°

Now, AB || CD and GE is a transversal.

∴ ∠FGE + ∠GED = 180° [Co-interior angles]

or ∠FGE + 126° = 180°

or ∠FGE = 180° –
126° = 54°

Thus, ∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.

**Ex 6.2 Class 9 Maths Question
4.**

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

[**Hint:**
Draw a line parallel to ST through point R.]

**Solution:**

Given, PQ || ST,

∠PQR = 110° and ∠RST = 130°^{ }Draw a line AB parallel to ST through R.

Now, ST || AB and SR is a transversal.

So, ∠RST + ∠SRB = 180° [Since, sum of the co-interior angles is 180°]

⇒ 130° + ∠SRB = 180°

⇒ ∠SRB = 180° – 130°

⇒ ∠SRB = 50° …(i)

Since, PQ
|| ST and AB || ST, so PQ || AB and then QR is a transversal.

So, ∠PQR + ∠QRA = 180° [Since, sum of the co-interior
angles is 180°]

⇒ 110° + ∠QRA = 180°

⇒ ∠QRA = 180° – 110°

⇒ ∠QRA = 70°
...(ii)

Now, ARB is
a line.

∴ ∠QRA + ∠QRS + ∠SRB = 180° [Angles on a line]

⇒ 70° + ∠QRS + 50° = 180°

⇒ 120° + ∠QRS = 180°

⇒ ∠QRS = 180° –
120°

⇒ ∠QRS = 60°

**Ex 6.2 Class 9 Maths Question
5.**

In the
given figure, if AB||CD, ∠APQ = 50°
and ∠PRD = 127°, find x and
y.

**Solution:**

In the given figure, AB || CD and PQ is a transversal.

∴ ∠APQ = ∠PQR [Alternate interior angles]

⇒ x = 50°

Also, AB ||
CD and PR is a transversal.

So, ∠APR = ∠PRD [Alternate interior angles]

⇒ 50° + y = 127° [∴ ∠APR = ∠APQ + ∠QPR = 50° + y]

⇒ y = 127° – 50°

⇒ y = 77°

Hence, x =
50° and y = 77°.

**Ex 6.2 Class 9 Maths Question
6.**

In the
given 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 AB || CD.

**Solution:**

Draw BE

⊥ PQ and CF ⊥ RS.

⇒ BE || CF

Also, ∠a = ∠b …(i) [∵ angle of incidence = angle of reflection]

and ∠x = ∠y …(ii) [∵ angle of
incidence = angle of reflection]

∴ ∠b = ∠x [Alternate interior angles]

⇒ 2∠b = 2∠x [Multiplying by 2 on both sides]

⇒ ∠b + ∠b = ∠x + ∠x

⇒ ∠a + ∠b = ∠x + ∠y [from eqs. (i) and (ii)]

⇒ ∠ABC = ∠DCB

which are alternate interior angles.

Hence, AB || CD.

**Hence proved.**

**NCERT Solutions for Maths Class 10**