NCERT Exemplar Problems Class 9 Maths – Lines and Angles

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.
Hence, it is a contradiction.
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:

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