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

## 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.

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:

Given, APQ = 50° and PRD = 127°
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 =                …(ii)              [ angle of incidence = angle of reflection]

Since, BE || CF and BC is a transversal.
b =                   [Alternate interior angles]
2b = 2x             [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.

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