# NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines And Angles

In this chapter, we provide NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 6 Lines And Angles for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 6 Lines And Angles pdf, free NCERT Exemplar Problems Solutions for Class 9 Maths Solutions Chapter 6 Lines And Angles book pdf download. Now you will get step by step solution to each question.

## NCERT Exemplar Class 9 Maths Solutions Lines And Angles

Exercise 6.1: Multiple Choice Questions (MCQs)

Question 1:
In figure, if AB || CD || EE, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(a) 85°             (b)135°                  (c)145°                   (d) 110°
Solution:
(c) Given, PQ || RS
∠PQC = ∠BRS = 60° [alternate exterior angles and ∠PQC = 60° (given)] and ∠DQR = ∠QRA = 25° [alternate interior angles]
[∠DQR = 25°, given]
∠QRS = ∠QRA + ∠ARS
= ∠QRA + (180° – ∠BRS) [linear pair axiom]
= 25° + 180° – 60°= 205° – 60°= 145°

Question 2:
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle             (b) an obtuse triangle
(c) an equilateral triangle         (d) a right triangle
Solution:
(d) Let the angles of a AABC be ∠A, ∠B and ∠C.
Given, ∠A = ∠B+∠C …(i)
InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of atriangle is 180°]…(ii)
From Eqs. (i) and (ii),
∠A+∠A = 180° => 2 ∠A = 180°
=> 180° /2
∠A = 90°
Hence, the triangle is a right triangle.

Question 3:
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is
(a) 37 ½°                  (b) 52 ½°               (c)72 ½°               (d) 75°
Solution:
Let one of interior angle be x°.
∴ Sum of two opposite interior angles = Exterior angle
∴  x° + x° = 105°
2x° = 105°
x° = 105°/2
x°=52 ½°
Hence, each angle of a triangle is 52 ½°.

Question 4:
If the angles of a triangle are in the ratio 5:3:7, then the triangle is
(a) an acute angled triangle
(b) an obtuse angled triangle
(c) a right angled triangle
(d) an isosceles triangle
Solution:
(a) Given, the ratio of angles of a triangle is 5 : 3 : 7.
Let angles of a triangle be ∠A,∠B and ∠C.
Then, ∠A = 5x, ∠B = 3x and ∠C = 7x
In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of all angles of a triangle is 180°]
5x + 3x + 7x = 180°
=> 15x = 180°
x = 180°/15= 12°
∠A = 5x = 5 x 12° = 60°
∠B = 3x= 3 x 12°= 36°
and ∠C =7x = 7 x 12° = 84°
Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Question 5:
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be
(a) 50°                (b) 65°                 (c) 145°                 (d) 155°
Solution:
(d) Let angles of a triangle  be ∠A, ∠B and ∠C. Question 6:
In the figure, POQ is a line. The value of x is

(a)20°              (b)25°            (c)30°               (d) 35°
Thinking Process
When two or more rays are initiated from a same point of a line, then the sum of all angles made between the rays and line at the same point is 180°.
Solution: Question 7:
In the figure, if OP || RS, ∠OPQ = 110° and ∠QRS = 130°, then ∠PQR is equal to

(a) 40°              (b) 50°                (c) 60°                          (d) 70°
Solution: Question 8:
Angles of a triangle are in the ratio 2:4:3. The smallest angle of the triangle is
(a) 60°             (b) 40°               (c) 80°                  (d) 20°
Thinking Process
Use the concept, the sum of all angles in a triangle is 180°. Further, simplify it and get the smallest angle.
Solution:
(b) Given, the ratio of angles of a triangle is 2 : 4 : 3.
Let the angles of a triangle be ∠A, ∠B and ∠C.
∠A = 2x, ∠B = 4x
∠C = 3x , ∠A+∠B+ ∠C= 180°
[sum of all the angles of a triangle is 180°]
2x + 4x + 3x = 180°
9x = 180°
x=180°/9 =20°
∠A=2x=2 x 20° = 40°
∠B = 4x = 4 x 20° = 80°
∠C = 3x = 3 x 20° = 60°
Hence, the smallest angle of a triangle is 40°.

Exercise 6.2: Very Short Answer Type Questions

Question 1:
For what value of x + y in figure will ABC be a line? Justify your answer.

Solution:
For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

Question 2:
Can a triangle have all angles less than 60°? Give reason for your answer.
Solution:
No, a triangle cannot have all angles less than 60°, because if all angles will be less than 60°, then their sum will not be equal to 180°. Hence, it will not be a triangle.

Question 3:
Can a triangle have two obtuse angles? Give reason for your answer.
Solution:
No, because if the triangle have two obtuse angles i.e., more than 90° angle, then the sum of all three angles of a triangle will not be equal to 180°.

Question 4:
How many triangles can be drawn having its angles as 45°, 64° and 72°? Give reason for your answer.
Solution:
None, the sum of given angles = 45° + 64° + 72° = 181° ≠ 180°.
Hence, we see that sum of all three angles is not equal to 180°. So, no triangle can be drawn with the given angles.

Question 5:
How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.
Solution:
Infinitely many triangles,
The sum of given angles = 53° + 64° + 63° = 180°
Here, we see that sum of all interior angles of triangle is 180°, so infinitely many triangles can be drawn.

Question 6:
In the figure, find the value of x for which the lines l and m are parallel.

Solution:
In the given figure, l || m and we know that, if a transversal intersects two parallel lines, then sum of interior angles on the same side of a transversal is supplementary. x + 44° = 180°
x = 180°-44°
=> x = 136° .

Question 7:
Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.
Solution:
No, because each of these will be a right angle only when they form a linear pair.

Question 8:
If one of the angles formed by two intersecting lines is a right angle, what can you say about the other three angles? Give reason for your answer.
Solution:
Let two intersecting lines l and m makes a one right angle, then it means that lines I and m are perpendicular each other. By using linear pair axiom aniom, other three angles will be a right angle.

Question 9:
In the figure, which of the two lines are parallel and why?

Solution:
In Fig. (i) sum of two interior angles 132° + 48° = 180° [∴ equal to 180°]
Here, we see that the sum of two interior angles on the same side of n is 180°, then they are the parallel lines.
In Fig. (ii), the sum of two interior angles 73° + 106° = 179° ≠ 180°. Here, we see that the sum of two interior angles on same side of r is not equal to 180°, then they are not the parallel lines.

Question 10:
Two lines l and m are perpendicular to the same line n. Are l and m perpendicular to each other? Give reason for your answer.
Solution:
No, since, lines l and m are perpendicular to the line n.
∠1 = ∠2 = 90° [∴ l ⊥ n and min]
It implies that these are corresponding angles.
Hence, l|| m.

Exercise 6.3: Short Answer Type Questions

Question 1:
In the figure, OD is the bisector of ∠AOC, OE is the bisector of ∠BOC and OD ⊥ OE. Show that the points A, 0 and B are collinear.

Thinking Process
For showing collinearity of A, O and B, we have to show that ∠AOB =180°.
Solution:
Given In the figure, OD ⊥ OE, OD and OE are the bisectors of ∠AOC and ∠BOC.
To show Points A, O and B are collinear i.e., AOB is a straight line.
Proof Since, OD and OE bisect angles ∠AOC and ∠BOC, respectively.
∠AOC =2 ∠DOC …(i)
and ∠COB = 2 ∠COE …(ii)
On adding Eqs. (i) and (ii), we get
∠AOC + ∠COB = 2 ∠DOC +2 ∠COE => ∠AOC +∠COB = 2(∠DOC +∠COE)
=> ∠AOC + ∠COB= 2 ∠DOE
=> ∠AOC+ ∠COB = 2 x 90° [∴ OD ⊥ OE]
=> ∠AOC + ∠COB = 180°
∴ ∠AOB = 180°
So, ∠AOC and ∠COB are forming linear pair.
Also, AOB is a straight line.
Hence, points A, O and B are collinear.

Question 2:
In the figure, ∠1 = 60° and ∠6 = 120°. Show that the lines m and n are parallel.

Solution:
Given In the figure ∠1 = 60° and ∠6 = 120°
To show m||n
Proof Since, ∠1 = 60° and ∠6 = 120°
Here, ∠1 = ∠3 [vertically opposite angles]
∠3 = ∠1 = 60°
Now, ∠3 + ∠6 = 60° + 120°
=> ∠3 + ∠6 = 180°
We know that, if the sum of two interior angles on same side of l is 180°, then lines are parallel.
Hence, m || n

Question 3:
AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ.

Solution:
Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively.
To prove AP|| BQ
Proof Since, l || m and t is transversal.
Therefore, ∠EAB = ∠ABH [alternate interior angles]

½ ∠EAB =½ ∠ABH    [dividing both sides by 2]
∠PAB =∠ABQ
[AP and BQ are the bisectors of ∠EAB and ∠ABH] Since, ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Question 4:
In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l ||m.

Solution:
Given, In the figure  AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF.
To show l || m
Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ
[alternate interior angles]
=> 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

So, alternate interior angles are equal.
We know that, if two alternate interior angles are equal, then lines are parallel. Hence, l || m.

Question 5:
In the figure, BA || ED and BC || EF. Show that ∠ABC = ∠DEF.

Solution:
Given BA || ED and BC || EF.
To show ∠ABC = ∠DEF.
Construction Draw a ray EP opposite to ray ED. Question 6:
In the figure, BA || ED and BC || EF. Show that ∠ABC + ∠DEF = 180°.

Solution:
Given BA || ED and BC || EF
To show, ∠ABC + ∠DEF = 180°
Construction Draw a ray PE opposite to ray EF. Question 7:
In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively. Find ∠APB.

Solution: Question 8:
A ΔABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that ∠BAL = ∠ACB.
Solution:
Given In ΔABC, ∠A = 90° and AL ⊥ BC
To prove ∠BAL = ∠ACB
Proof In ΔABC and ΔLAC, ∠BAC = ∠ALC  [each 90°] …(i)

and ∠ABC = ∠ABL [common angle] …(ii)

On adding Eqs. (i) and (ii), we get
∠BAC + ∠ABC = ∠ALC + ∠ABL …(iii)
Again, in ΔABC,
∠BAC + ∠ACB + ∠ABC = 180°
[sum of all angles of a triangle is 180°] =>∠BAC+∠ABC = 1 80°-∠ACB …(iv)
In  ΔABL,
∠ABL + ∠ALB + ∠BAL = 180°
[sum of all angles of a triangle is 180°] => ∠ABL+ ∠ALC = 180° – ∠BAL [∴ ∠ALC = ∠ALB= 90°] …(v)
On substituting the value from Eqs. (iv) and (v) in Eq. (iii), we get 180° – ∠ACS = 180° – ∠SAL
=> ∠ACB = ∠BAL
Hence proved.

Question 9:
Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.
Solution:
Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n.
i.e.,  p ⊥ m,  p ⊥ n, q ⊥ m, q ⊥ n
To prove p||g
Proof Since, m || n and p is perpendicular to m and n. So, sum of two interior angles is supplementary.
We know that, if a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.
Hence, p||g.

Exercise 6.4: Long Answer Type Questions

Question 1:
If two lines intersect prove that the vertically opposite angles are equal
Solution:
Given Two lines AB and CD intersect at point O. Question 2:
Bisectors of interior ∠B and exterior ∠ACD of a  ΔABC intersect at the point T. Prove that ∠BTC = ½ ∠BAC.
Thinking Process
For obtaining the interior required result use the property that the exterior angle of a triangle is equal to the sum of the two opposite angles of triangle.

Solution:
Given In AABC, produce SC to D and the bisectors of  ∠ABC and ∠ACD meet at point T. To prove  ∠BTC = ½ ∠BAC Question 3:
A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.
Solution:
Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and  ∠CQP, respectively. Question 4:
Prove that through a given point, we can draw only one perpendicular to a given line.
Solution:
Given Consider a line l and a point P. Question 5:
Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other.
Solution:
Given Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D.

To prove Two lines n and p intersecting at a point.
Proof Suppose we consider lines n and p are not intersecting, then it means they are parallel to each other i.e., n || p …(i)
Since, lines n and pare perpendicular to m and l, respectively.
But from Eq. (i) n || p it implies that l || m.
Thus, our assumption is wrong.
Therefore, lines n and p intersect at a point.

Question 6:
Prove that a triangle must have atleast two acute angles.
Solution:
Given ΔABC is a triangle.
To prove ΔABC must have two acute angles
Proof Let us consider the following cases
Case I When two angles are 90°.
Suppose two angles are ∠B = 90° and ∠C = 90° Question 7:
In the given figure, ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR.
Prove that ∠APM = ½(∠Q – ∠R).

Solution: All Chapter NCERT Exemplar Problems Solutions For Class 9 maths

—————————————————————————–

All Subject NCERT Exemplar Problems Solutions For Class 9

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share ncertexemplar.com to your friends.