ncert class 9 maths chapter 6 solution, Lines and Angles Exercise 6.2 involve complete answers for each question in the exercise 6.2. The solutions provide students a strategic methods to prepare for their exam. Class 9 Maths Chapter 6 Lines and Angles exercise 6.2 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems.NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way.
1. In Fig. 6.28, find the values of x and y and then show that AB || CD.
Solution 1:
It can be observed that,
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.
2. In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7,find x.
Solution 2:
It is given that AB || CD and CD || EF
AB || CD || EF (Lines parallel to the same line are parallel to each other)
It can be observed that
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)
z + y = 180º [Using Equation (1)]
7a + 3a = 180º
10a = 180º
a = 18º
∴ x = 7a = 7 × 18º = 126º
3. In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.
Solution 3:
It is given that,
AB || CD and EF || CD
∠GED = 126º
∠GEF + ∠FED = 126º
∠GEF + 90º = 126º
∠GEF = 36º
As ∠AGE and ∠GED are alternate interior angles.
∠AGE = ∠GED = 126º
However, ∠AGE + ∠FGE = 180º (Linear pair)
126º + ∠FGE = 180º
∠FGE = 180º − 126º = 54º
Hence, ∠AGE = 126º, ∠GEF = 36º, ∠FGE = 54º
4. In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.
[Hint : Draw a line parallel to ST through point R.]
Solution 4:
Let us draw a line XY parallel to ST and passing through point R.
∠PQR + ∠QRX = 180º (Co-interior angles on the same side of transversal QR)
110º + ∠QRX = 180º
∠QRX = 70º
Also,
∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)
130º + ∠SRY = 180º
∠SRY = 50º
XY is a straight line. RQ and RS stand on it.
∠QRX + ∠QRS + ∠SRY = 180º
70º + ∠QRS + 50º = 180º
∠QRS = 180º − 120º = 60º
5. In Fig. 6.32, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y.
Solution 5:
∠APR = ∠PRD (Alternate interior angles)
50º + y = 127º
y = 127º − 50º
y = 77º
Also,
∠APQ = ∠PQR (Alternate interior angles)
50º = x
Therefore, x = 50º and y = 77º
6. In Fig. 6.33, 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.
Let us draw BM ⊥PQ and CN ⊥RS.
As PQ || RS,
Therefore, BM || CN
Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.
∠2 = ∠3 (Alternate interior angles)
However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)
∠1 = ∠2 = ∠3 = ∠4
Also,
∠1 + ∠2 = ∠3 + ∠4
∠ABC = ∠DCB
However, these are alternate interior angles.
therefore, AB || CD
