Class 9 Exam  >  Class 9 Notes  >  Mathematics Olympiad for Class 9  >  NCERT Solutions: Lines & Angles (Exercise 6.2)

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)

Q1. In the following Figure, if AB CD, CD EF and y : z = 3 : 7, find x.

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)

Ans: It is known that ABCD and CDEF
As the angles on the same side of a transversal line sums up to 180°,
x + y = 180° —–(i)
Also,
O = z (Since they are corresponding angles)
and, y +O = 180° (Since they are a linear pair)
So, y+z = 180°
Now, let y = 3w and hence, z = 7w (As y : z = 3 : 7)
∴ 3w+7w = 180°
Or, 10 w = 180°
So, w = 18°
Now, y = 3×18° = 54°
and, z = 7×18° = 126°
Now, angle x can be calculated from equation (i)
x+y = 180°
Or, x+54° = 180°
∴ x = 126°


Q2. In the following figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)Ans: Since AB || CD, GE is a transversal.
It is given that ∠GED = 126°
So, ∠GED = ∠AGE = 126° (As they are alternate interior angles)
Also,
∠GED = ∠GEF +∠FED
As EF⊥ CD, ∠FED = 90°
∴ ∠GED = ∠GEF+90°
Or, ∠GEF = 126° – 90° = 36°
Again, ∠FGE +∠GED = 180° (Transversal)
Putting the value of ∠GED = 126°, we get
∠FGE = 54°
So,
∠AGE = 126°
∠GEF = 36° and
∠FGE = 54°


Q3. In the following figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint : Draw a line parallel to ST through point R.]

NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)

Ans: First, construct a line XY parallel to PQ.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)
We know that the angles on the same side of transversal is equal to 180°.
So, ∠PQR+∠QRX = 180°
Or, ∠QRX = 180°-110°
∴ ∠QRX = 70°
Similarly,
∠RST +∠SRY = 180°
Or, ∠SRY = 180°- 130°
∴ ∠SRY = 50°
Now, for the linear pairs on the line XY-
∠QRX+∠QRS+∠SRY = 180°
Putting their respective values, we get
∠QRS = 180° – 70° – 50°
Hence, ∠QRS = 60°


Q4. In the following figure, if AB || CD, ∠ APQ = 50° and ∠ PRD = 127°, find x and y
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)Ans: From the diagram,
∠APQ = ∠PQR (Alternate interior angles)
Now, putting the value of ∠APQ = 50° and ∠PQR = x, we get
x = 50°
Also,
∠APR = ∠PRD (Alternate interior angles)
Or, ∠APR = 127° (As it is given that ∠PRD = 127°)
We know that
∠APR = ∠APQ+∠QPR
Now, putting values of ∠QPR = y and ∠APR = 127°, we get
127° = 50°+ y
Or, y = 77°
Thus, the values of x and y are calculated as:
x = 50° and y = 77°


Q5. In the following 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.
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)
Ans: First, draw two lines, BE and CF, such that BE ⊥ PQ and CF ⊥ RS.
Now, since PQ || RS,
So, BE || CF
NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)

We know that,
Angle of incidence = Angle of reflection (By the law of reflection)
So,
∠1 = ∠2 and
∠3 = ∠4
We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C
So, ∠2 = ∠3 (As they are alternate interior angles)
Now, ∠1 +∠2 = ∠3 +∠4
Or, ∠ABC = ∠DCB
So, AB || CD (alternate interior angles are equal)

The document NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2) is a part of the Class 9 Course Mathematics Olympiad for Class 9.
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FAQs on NCERT Solutions for Class 9 Maths Chapter 6 - Lines and Angles (Exercise 6.2)

1. What are the different types of angles?
Ans. There are several types of angles, including acute angles (less than 90 degrees), obtuse angles (greater than 90 degrees), right angles (exactly 90 degrees), straight angles (exactly 180 degrees), and reflex angles (greater than 180 degrees).
2. How can I identify parallel lines?
Ans. Parallel lines are lines that never intersect each other. To identify parallel lines, you can look for pairs of lines that have the same slope. If the slopes of two lines are equal, then they are parallel.
3. What is the sum of the angles in a triangle?
Ans. The sum of the angles in a triangle is always 180 degrees. This property is known as the Angle Sum Property of a Triangle.
4. How can I find the measure of an angle using a protractor?
Ans. To find the measure of an angle using a protractor, place the center point of the protractor on the vertex of the angle. Then, align one side of the angle with the baseline of the protractor. The measure of the angle can be read where the other side of the angle intersects the protractor scale.
5. Can two lines intersect at more than one point?
Ans. No, two lines can only intersect at a single point. If two lines intersect at more than one point, then they are not considered to be lines but rather the same line. This is known as the Unique Intersection Property of Lines.
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