Chapter 6 Lines and Angles | Class 9 NCERT CBSE Mathematics Solution | NCERT Maths Textbook Exercise 6.2

 

Class IX NCERT CBSE Mathematics Solutions
Lines and Angles
NCERT Mathematics Textbook Exercise 6.1 Solved
(Page 103)
Q1: In the given figure, find the values of x and y and then show that AB || CD.
Extrascore NCERT CBSE Class 9, Math solutions
Ans:
In the above figure,
50º + x = 180º (Linear pair)
x = 130º … (1)
Also, y = 130º (Vertically opposite angles)
As x and y are alternate interior angles for lines AB and CD and also measures of these angles are equal to each other, therefore, line AB || CD.
Q2: In the given figure, if AB || CD, CD || EF and y: z = 3: 7, find x.
Extrascore NCERT CBSE Class 9, Math solutions
Ans:
Given that AB || CD and CD || EF
or, AB || CD || EF (Lines parallel to the same line are parallel to each other)
Also, x = z (Alternate interior angles) . . . . . . . (1)
It is given that y : z = 3 : 7
Let the common ratio between y and z be a.
y = 3a and z = 7a
Also, x + y = 180º (Co-interior angles on the same side of the transversal)
or, z + y = 180º [from equation (1)]
or, 7a + 3a = 180º
or, 10a = 180º
or, a = 18º
so, x = 7a = 7 × 18º = 126º
Q3: In the given figure, If AB || CD, EF CD and GED = 126º, find AGE, GEF and FGE.
Extrascore NCERT CBSE Class 9, Math solutions
Ans:
It is given that, AB || CD, EF ⊥ CD and ∠GED = 126º
From the figure, ∠GED = ∠GEF + ∠FED = 126º
or, ∠GEF + 90º = 126º
or, ∠GEF = 36º
(∠AGE and ∠GED are alternate interior angles.)
or, ∠AGE = ∠GED = 126º
However, ∠AGE + ∠FGE = 180º (Linear pair)
or, 126º + ∠FGE = 180º
or, ∠FGE = 180º − 126º = 54º
Therefore, ∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º
Q4: In the given figure, if PQ || ST, PQR = 110º and RST = 130º, find QRS.
[Hint: Draw a line parallel to ST through point R.]
Extrascore NCERT CBSE Class 9, Math solutions













Let us draw a line XY || ST and passing through point R.
we can write, ∠PQR + ∠QRX = 180º 
(Co-interior angles on the same side of transversal QR)
or, 110º + ∠QRX = 180º
or, ∠QRX = 70º
Also, ∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)
or, 130º + ∠SRY = 180º
or, ∠SRY = 50º
XY is a straight line and RQ and RS intersect XY at the point R.
∴ ∠QRX + ∠QRS + ∠SRY = 180º
or, 70º + ∠QRS + 50º = 180º
or, ∠QRS = 180º − 120º = 60º
Q5: In the given figure, if AB || CD, APQ = 50º and PRD = 127º, find x and y.
Extrascore NCERT CBSE Class 9, Math solutions
Ans:
In the above figure, 
∠APR = ∠PRD (Alternate interior angles)
or, 50º + y = 127º
or, y = 127º − 50º
or, y = 77º
Also, ∠APQ = ∠PQR (Alternate interior angles)
x = 50º and y = 77º
Q6: 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. 
Extrascore NCERT CBSE Class 9, Math solutions













Let us draw BM ⊥ PQ and CN ⊥ RS.
Since, PQ || RS
Therefore, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC intersects these lines at B and C respectively.
So, ∠2 = ∠3 (Alternate interior angles)
However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
So, ∠1 = ∠2 = ∠3 = ∠4
Also, ∠1 + ∠2 = ∠3 + ∠4
or, ∠ABC = ∠DCB (but, these are alternate interior angles)
∴ AB || CD

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