NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles 2021

NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles

Edited By Safeer PP | Updated on Aug 31, 2022 11:56 AM IST

NCERT exemplar Class 9 Maths solutions chapter 6 provide students with detailed answers for chapter 6- Lines and Angles. These problems and their solutions are devised to understand the chapter better and extremely useful in understanding Lines and Angles concepts. The NCERT exemplar Class 9 Maths chapter 6 solutions are intricate and helpful in pinpointing the critical understanding of concepts and to solve questions in NCERT Class 9 Maths Book. The exemplar questions and answers are based on the recommended CBSE Class 9 Syllabus. These Class 9 Maths NCERT exemplar solutions chapter 6 Lines and Angles solutions are a rich source of reference material equipped with numerous practice problems of lines and angles.

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Question:1

Write the correct answer in each of the following:
$\text{In Fig. 6.1, if }AB \parallel CD \parallel EF, PQ \parallel RS,$ $\angle RQD = 25^{\circ}$ $\text{ and }\angle CQP = 60^{\circ}$ $\text{ and }\angle CQP = 60^{\circ} \text{, then}$ $\angle QRS \text{ is equal to }$

$(A)\ 85^{\circ}$

$(B)\ 135^{\circ}$

$(C)\ 145^{\circ}$
$(D)\ 110^{\circ}$

Answer:

Here in this question, it is given,
$PQ \parallel RS$
$\angle PQC = \angle BRQ = 60^{\circ}$ $\text{(Alternate Exterior angles)}$
$\angle PQC = 60^{\circ}$
$\angle DQR = \angle QRA= 25^{\circ}$ $\text{(Alternate interior angles)}$
$\angle QRS = \angle QRA + \angle ARS$
$= \angle QRA + (180^{\circ} - \angle BRS)$ $\text{(Linear pair angle)}$
$= 25^{\circ} + 180^{\circ} - 60^{\circ} = 205^{\circ} - 60^{\circ} = 145^{\circ}$

Hence, option C is the right answer.

Question:2

Write the correct answer in each of the following:
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

Answer: [D] a right triangle

In a $\triangle ABC$

$\angle A = \angle B +\angle C .....(i)$

$\text{Now sum of all angles =}180^{\circ}$

$\angle A +\angle B + \angle C = 180^{\circ}$
$\angle A+\angle A=180^{\circ}$ $\text{(from i)}$
$\angle A=90^{\circ}$

Thus triangle ABC is a right-angled triangle.

Hence Option (D) is the correct answer.

Question:3

Write the correct answer in each of the following:
An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is
$(A)$ $34\frac{1}{2}^{\circ}$
$(B)$ $52\frac{1}{2}^{\circ}$
$(C)$ $72\frac{1}{2}^{\circ}$
$(D)$ $75^{\circ}$

Answer:

Given ® Exterior Angle = $105^{\circ}$

Let the two equal interior angle be $x^{\circ}$ .

Sum of opposite interior angles of a triangle = Exterior angle of the triangle.

$x^{\circ}+x^{\circ}=105^{\circ}$

$2x^{\circ} = 105^{\circ}$

$x^{\circ}=52\frac{1}{2}^{\circ}$

So, option (B) is the correct Answer

Question:4

Write the correct answer in each of the following: The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is
(A) an acute angled triangle
(B) an obtuse angled triangle
(C) a right triangle
(D) an isosceles triangle

Answer: (A) an acute angled triangle

Given: The ratio of angles of triangles is 5 : 3 : 7

Let angles of the triangle be $\angle A,\angle B$ $and$ $\angle C$

$\angle A : \angle B : \angle C$

$5: 3 : 7$ $5x: 3x : 7x$ $\text{ here x = constant }$

Then, $\angle A = 5x$

$\angle B = 3x$

$\angle C = 7x$

In $\triangle ABC$

$\angle A + \angle B + \angle C = 180^{\circ}$

$5x + 3x + 7x = 180^{\circ}$

$15x = 180^{\circ}$

$x=\frac{180}{18}=12^{\circ}$

$\angle A = 5x = 5\times 12^{\circ}= 60^{\circ}$

$\angle B = 3x = 3 \times 12^{\circ} = 36^{\circ}$

$\angle C = 7x = 7 \times 12^{\circ} = 84^{\circ}$

All angles are less than 90° hence acute.

Question:5

Write the correct answer in each of the following:
If one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles can be

(A) $50^{\circ}$

(B) $65^{\circ}$

(D) $155^{\circ}$

Answer: (D) $155^{\circ}$

In $\triangle ABC$

$\angle A +\angle B + \angle C = 180^{\circ}$

[Sum of all interiors angles of triangle is $180^{\circ}$ ]

$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B + \frac{1}{2}\angle C =$ $\frac{180^{\circ}}{2}$ $=90^{\circ}$

$\Rightarrow \frac{1}{2} \angle B+ \frac{1}{2}\angle C=90^{\circ}-\frac{1}{2}\angle A................(1)$

$\text{Since, in}\triangle BOC$

$\text{As BO and OC are the angle bisectors of }\angle ABC$ $\text{and }\angle BCA$ ,

$\frac{\angle B}{2}+ \frac{\angle C}{2}+\angle BOC=180^{\circ}..................(2)$

Put value of equation (1) in equation (2)

$90^{\circ} -\frac{1}{2}\angle A+\angle BOC =180^{\circ}$
$\angle BOC = 180^{\circ} - 90^{\circ} + \frac{1}{2}\angle A$

$\angle BOC = 90^{\circ} + \frac{1}{2}\angle A$

$\therefore \angle A=130^{\circ}$

$\angle BOC= 90^{\circ}+\frac{1}{2}\times 130^{\circ}$

$\angle BOC= 90^{\circ} + 65^{\circ}$

$\angle BOC= 155^{\circ}$

Question:6

Write the correct answer in each of the following:
In Fig. 6.2, POQ is a line. The value of x is

$(A)$ $20^{\circ}$
$(B)$ $25^{\circ}$
$(C)$ $30^{\circ}$
$(D)$ $35^{\circ}$

Answer:

Given POQ is a line

$\angle POQ = 180^{\circ}$ (angle on a straight line)

$40^{\circ} + 4x + 3x = 180^{\circ}$

$7x = 140^{\circ}$

$x = 20^{\circ}$

Question:7

Write the correct answer in each of the following:
$\text{ In Fig. 6.3, if }OP\parallel RS$ , $\angle OPQ = 110^{\circ}\text{and }$ $\angle QRS = 130^{\circ}\text{, then }$ $\angle PQR\text{ is equal to }$

$(A)\ 40^{\circ}$
$(B)$ $50^{\circ}$
$(C)$ $60^{\circ}$
$(D)$ $70^{\circ}$

Answer:

$OP \parallel RS$

$\angle RWV+\angle SRW=180^{\circ}$ $\text{ (sum of co- interior angles) }$ $130^{\circ} + \angle RWV =180^{\circ}$

$\angle RWV = 50^{\circ}$
$\angle PWQ = \angle RWV = 50^{\circ}\text{(vertically opposite angles are equal) }$

Also, for line OP

$\angle OPQ +\angle WPQ = 180^{\circ}\text{(linear pair)}$

$\angle WPQ = 180^{\circ} - \angle OPQ$

$= 180^{\circ} -110^{\circ}$

$\angle WPQ = 70^{\circ}$

Now, we know that the sum of angles of a triangle is 180⁰

$\text{in }\triangle PQW$ ,

$\angle WPQ + \angle PQW +\angle PWQ = 180^{\circ}$

$\angle PQW= 180^{\circ} - \angle WPQ -\angle PWQ = 180^{\circ} - 70^{\circ} - 50^{\circ}$

$\angle PQR = 180^{\circ} - 120^{\circ}$

$\angle PQR = 60^{\circ}$

Hence option C is the correct answer

Question:8

Write the correct answer in each of the following: Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is
$(A)$ $60^{\circ}$
$(B)$ $40^{\circ}$
$(C)$ $80^{\circ}$
$(D)$ $20^{\circ}$

Answer:

$\text{Let the angles of the triangle be}$ $\angle A ,\angle B,\angle C$ in the ratio 2:4:3

$\angle A = 2x$
$\angle B = 4x$
$\angle C = 3x$
We know that some of the angle of the triangle is 180⁰
9x=180
x=20
Angle A=40, B=80, C=60

Hence option (B) is correct

Question:1

For what value of x+y in Fig. 6.4 will ABC be a line? Justify your answer.

Answer:

Here x & y are two adjacent angles and for straight-line x+y=180⁰

Justification: As we know that if the sum of two adjacent angles is 180⁰ it represents the straight line.

$\angle CBD= x,$ $\angle ABD =y$

$\text{To form ABC; a straight line the sum of}$ $\angle ABD+\angle CBD=180^{\circ}$

$x + y = 180^{\circ}$

Question:2

Can a triangle have all angles less than 600? Give reason for your answer.

Answer:

In a triangle, the sum of interior angles is always equal to 180⁰

So it is not possible to have all angles less than 60⁰because their sum will not be equal to 180⁰

Hence, the answer is No

Question:3

Can a triangle have two obtuse angles? Give reason for your answer

Answer: No

No, a triangle cannot have two obtuse angles
Reason: the obtuse angle is greater than 90⁰
Let us consider the smallest obtuse angle, i.e., 91⁰
Let the two obtuse angles be equal to 91^o, so the sum of these two obtuse angles is equal to 182⁰
But the sum of interior angles of a triangle is always equal to 180⁰
So the given condition is not possible.

Hence, we cannot create a triangle with two obtuse angles

Question:4

How many triangles can be drawn having its angles as 450, 640 and 720? Give reason for your answer.

Answer: Zero

The three angles given are 45⁰, 64⁰ and 72⁰

Now, the sum of these angles = 45⁰+64⁰+72⁰

= 181^o

Sum of angles of a triangle cannot be more than 180⁰.

Hence, the triangle is not possible
So the answer is zero.

Question:5

How many triangles can be drawn having its angles as 530, 640 and 630? Give reason for your answer.

Answer: infinitely many

Infinitely many triangles can be drawn having its angles as 53⁰, 64⁰ and 63⁰

The sum of given angles = 53⁰ + 64⁰⁺ 63⁰ = 180⁰

Here, we can see that sum of all interior angles of triangles is 180⁰, so infinitely many triangles can be drawn depending on the lengths of its sides.

So, infinitely many triangles can be drawn from the given angles.

Question:6

In Fig. 6.5, find the value of x for which the lines l and m are parallel.

Answer: $\angle x=136^{\circ}$

In the given figure, the lines l and m are parallel.

Line n is a transversal line.

So, $\angle x$ and 44^o are co-interior angles.

We know that the sum of co-interior angles is 180^o.

Hence $\angle x + 44^{\circ} = 180^{\circ}$

$\angle x = 136^{\circ}$ is the correct answer

Question:7

Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.

Answer: No

Given that two adjacent angles are equal.

No it is not necessary that each of these angles will be a right angle.
Let us see why.
Adjacent angles: Two angles are said to be adjacent only when they have common vertex, a common side but they do not overlap.
The following figure shows adjacent angles ( $\angle 1$ and $\angle 2$ ), (<DBC, $\angle CBA$ )

But for the case that both adjacent angles are right angles, they should form a linear pair or we can say that they should lie on the same line, but such condition is not given to us.

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.

Answer: All the other angles should be at right angles.

Let AB and CD be two intersecting lines which intersect at point O

If one of the angle is formed by two intersecting line is a right angle,
$\text{Then let}$ $\angle AOD=90^{\circ}$
$\angle AOD=\angle BOC=90^{\circ}$ $\text{(vertically opposite angles)}$
$\angle AOD,\angle BOD$ $\text{form linear pair}$

$\angle AOD+\angle BOD=180^{\circ}$
$\text{Hence,}$ $\angle BOD=90^{\circ}$
$\angle BOD=\angle AOC=90^{\circ}$ $\text{ (vertically opposite angles)}$

Hence all the other angles should be at right angles.

Question:9

In Fig.6.6, which of the two lines are parallel and why?

Answer: l and m are parallel

Consider l and m, now n is the transversal
$132^{\circ} + 48^{\circ} = 180^{\circ}$ $(sum of co-interior angles = 180^{\circ})$
Hence the lines are parallel
Consider p and q, now r is the transversal
$106^{\circ} + 73^{\circ} = 179^{\circ}\neq 180^{\circ}$ $(sum of co-interior angles \neq 180^{\circ})$
Hence the lines are not parallel
Therefore, l and m are parallel

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.

Answer: No.

The two lines l and m are perpendicular to the same line n.
We can have 3 possible scenarios:
1. The lines l and m may lie on the same side of n

$\angle lAn=\angle mCn=90^{\circ}$
As we can see corresponding angles are equal, so l is parallel to m
l and m are not perpendicular to each other
2. The lines l and m may lie on the opposite side of n
Extend lA to point B and mC to point D

$\angle lAC=\angle mCA=90^{\circ}$
As we can see that interior opposite angles are equal, so l is parallel to m
l and m are not perpendicular to each other
3. The lines l and m are the same and overlap each other.
Again, l and m are not perpendicular to each other

Hence, the lines l and m are not perpendicular.

Question:1

In Fig. 6.9, OD is the bisector of , OE is the bisector of and Show that the points A, O and B are collinear.

Answer:

Given in the figure $OD \perp OE.$ , OD and OE are bisectors of $\angle AOC$ and $\angle BOC$ respectively
To show: Points A, O & B are collinear, i.e., AOB is a straight line
Proof: Since OD and OE bisect angles $\angle AOC$ and $\angle BOC$ respectively
$\angle AOC = 2 \angle DOC$ … (i)
And $\angle COB = 2 \angle COE$ … (ii)
On adding equation (i) and (ii) we get
$\angle AOC + \angle COB = 2 \angle DOC + 2 \angle COE$
$\angle AOC + \angle COB = 2 (\angle DOC + \angle COE)$
$\angle AOC + \angle COB = 2 \angle DOE$
$\angle AOC + \angle COB = 2 \times 90^{\circ}$ (Given $OD \perp OE$ )
$\angle AOC + \angle COB = 180^{\circ}$
$\angle AOB = 180^{\circ}$
So, $\angle AOC$ & $\angle COB$ are forming a linear pair.
AOB is a straight line.
Therefore, points A, O and B are collinear.
Hence proved

Question:2

$\text{In Fig. 6.10,}\ \angle 1 = 60^{\circ}$ $\text{ and}$ $\angle 6 = 120^{\circ}$ . Show that the lines m and n are parallel.

Answer:

Given: The figure $\angle 1 = 60^{\circ}$ and $\angle 6 = 120^{\circ}$
To show : $m \parallel n$
Proof: $\angle 1 = 60^{\circ}$ and $\angle 6 = 120^{\circ}$
Here, $\angle 1 = 60^{\circ}$ and $\angle 6 = 120^{\circ}$
Here $\angle 1 = \angle 3$ (Vertically opposite angles)
$\angle 3 = \angle 1 = 60^{\circ}$
Now, $\angle 3 + \angle 6 = 60^{\circ}+ 120^{\circ}$
$\angle 3 + \angle 6 = 180^{\circ}$
We know if the sum of two interior angles on the same side is $180^{\circ}$ then the lines are parallel
Hence proved, $m \parallel n$ .

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 (Fig. 6.11). Show that AP parallel to BQ

Answer:

Given: In the figure, $l \parallel m$ , AP and BQ are the Bisectors of $\angle EAB$ and $\angle ABH$
To prove: $AP \parallel BQ$
Proof: Since $l \parallel m$ and t is a transversal therefore
$\angle EAB = \angle ABH$ (alternate interior angles)

$\frac{1}{2} \angle EAB = \frac{1}{2}\angle ABH$ (Divide both sides by 2)
$\angle PAB = \angle ABQ$ (AP & BQ are the bisectors of $\angle EAB$ & $\angle ABH$ )
Now consider, two lines AP and BQ with transversal AB
$\angle PAB$ and $\angle ABQ$ are alternate interior angles and these are equal.
Hence, $AP \parallel BQ$
Hence proved

Question:4

If in Figure bisectors AP and BQ of the alternate interior angles are parallel, then show that .

Answer:

In the figure $AP \parallel BQ$ , AP and BQ are the bisectors of the alternate interior angles $\angle CAB$ and $\angle ABF$ .

To show : $l \parallel m$
Proof: Since $AP \parallel BQ$ and t is transversal therefore
$\angle PAB = \angle ABQ$ (Alternate Interior Angles)
$\Rightarrow 2 \angle PAB = 2\angle ABQ$ (Multiplying both sides by 2)
Now, AP and BQ are the bisectors of alternate interior angle $\angle CAB$ and $\angle ABF$
$2 \angle PAB = \angle CAB$
$2\angle ABQ = \angle ABF$
So,
$\angle CAB = \angle ABF$
Now consider lines l and m
$\angle CAB = \angle ABF$ (alternate interior angles are equal)
Hence $l \parallel m$

Hence proved

Question:5

In Fig. 6.12, $BA \parallel ED and BC \parallel EF$ . Show that $\angle ABC = \angle DEF$ [Hint: Produce DE to intersect BC at P (say)].

Answer:

Produce DE to intersect BC at P

Now, $EF \parallel BC$ and DP is the transversal
$\angle DEF = \angle DPC$ (corresponding angles) … (i)
Now $AB \parallel DP$ and BC is transversal,
$\angle DPC = \angle ABC$ (corresponding angles) … (ii)
From (i) and (ii) we get
$\angle ABC = \angle DEF$

Hence, Proved

Question:6

In Fig. 6.13, $BA \parallel ED$ and $BC \parallel EF$ . Show that $\angle ABC +\angle DEF = 180^{\circ}$

Answer:

Given
$BA \parallel ED$ and $BC \parallel EF$
To show $\angle ABC + \angle DEF = 180^{\circ}$
Construction:
Extend EF to point P on AB

Proof: In figure, $BC \parallel EF$ , so $BC \parallel PF$
$\because \angle EPB + \angle PBC = 180^{\circ}$ (Sum of co-interior angles is $180^{\circ}$ ) …(i)
Now, $AB \parallel ED$ and PE is the transversal line.
$\angle EPB = \angle DEF$ (Corresponding angle) …(ii)
From Equations (i) and (ii)
$\angle DEF + \angle PBC = 180^{\circ}$
$\angle ABC + \angle DEF = 180^{\circ} (\because \angle PBC = \angle ABC)$

Hence proved

Question:7

In Fig. 6.14, $DE \parallel QR$ and AP and BP are bisectors of $\angle EAB$ and $\angle RBA$ , respectively. Find $\angle APB.$

Answer: $\angle APB = 90^{\circ}$

$DE \parallel QR$ and AP and PB are the bisectors of $\angle EAB$ and $\angle RBA$ ,
We know that the interior angles on the same side of the transversal are supplementary
So,
$\angle EAB + \angle RBA = 180^{\circ}$
$\frac{1}{2} \angle EAB + \frac{1}{2}\angle RBA =\frac{1}{2} (180^{\circ})$
$\frac{1}{2}\angle EAB+\frac{1}{2}\angle RBA=90^{\circ}$ …(i)
AP and BP are the bisectors of $\angle EAB$ and $\angle RBA$ respectively.
$\angle BAP=\frac{1}{2}\angle EAB$ …(ii)
$\angle ABP = \frac{1}{2} \angle RBA$ …(iii)
On adding equation (ii) and (iii) we get
$\angle BAP + \angle ABP = \frac{1}{2} \angle EAB +\frac{1}{2} \angle RBA$
From equation (i)
$\angle BAP + \angle ABP = 90^{\circ}$
In $\triangle APB$ ,
$\angle BAP + \angle ABP + \angle APB = 180^{\circ}$
$90^{\circ}+ \angle APB = 180^{\circ}$
$\angle APB = 180^{\circ} - 90^{\circ}= 90^{\circ}$

Hence $\angle APB = 90^{\circ}$

Question:8

The angles of a triangle are in the ratio 2 : 3 : 4. Find the value of each angle. What type of triangle is it?

Answer:

Angles of the triangle are in the ratio- 2 : 3 : 4
Let the angles are 2x, 3x, 4x then:
$2x + 3x + 4x = 180^{\circ}$ (angle sum property)
$9x = 180^{\circ}$
$x = 20^{\circ}$
Then the angles of the triangle are:
$2x = 40^{\circ}$
$3x = 60^{\circ}$
$4x = 80^{\circ}$
This triangle is a scalene triangle as all the angles are of different measure.

Question:9

A triangle ABC is right angled at A. L is a point on BC such that $AL\perp BC$ . Prove that $\angle BAL = \angle ACB$ .

Answer:

$AL\perp BC$ $\angle BAL = \angle ACB$

Proof: In $\triangle ABC$ and $\triangle LAB$ ,
$\angle BAC = \angle ALB$ (each $90^{\circ}$ ) (i)
And $\angle ABC = \angle ABL$ (Common angle) (ii)
In $\triangle ABC$ ,
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$ (angle sum property)
$\angle ABC + \angle ACB = 180^{\circ} -90^{\circ}$ (from i)
$\angle ACB = 90^{\circ} - \angle ABC$
In $\triangle ABL$ ,
$\angle BAL + \angle ALB + \angle ABL = 180^{\circ}$ (angle sum property)
$\angle BAL + \angle ABL = 180^{\circ} - 90^{\circ}$ (from i)
$\angle BAL = 90^{\circ} - \angle ABL$
$\angle BAL = 90^{\circ} - \angle ABC$ (from ii)
Hence, $\angle ACB = 90^{\circ} -\angle ABC = \angle BAL$

Hence proved

Question:10

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer:

Given: Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n, i.e., $p \perp m, p \perp n, q \perp m, q \perp n$ .
To prove
$p \parallel q$

Proof:

Since m||n and p is perpendicular to m and n.
$\angle 5 = \angle 6 = \angle 7 = \angle 8 = \angle 11 = \angle 12 = 90^{\circ}$
Similarly, q is perpendicular to m and n.
$�\\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 9 = \angle 10 = 90^{\circ}$
Now for lines p and q, m is the transversal
$\angle 1 = \angle 2 = \angle 3 = \angle 4 = \angle 5 = \angle 6 = \angle 7 = \angle 8 = 90^{\circ}$
So we can see that all the conditions are fulfilled for the lines to be parallel, i.e., Corresponding angles are equal, sum of cointerior angles is 180^o, alternate angles are equal.

Hence, $p \parallel q$

Hence proved

Question:1

If two lines intersect, prove that the vertically opposite angles are equal.

Answer:

It is given that if two lines intersect, the vertically opposite angles are equal.
Proof:

Now let AB and CD be two lines intersecting at point O.

From the figure, we have two pairs of vertically opposite angles namely:

(i) $\angle AOC$ and $\angle BOD$

(ii) $\angle AOD$ and $\angle BOC$

Now we have to prove that $\angle AOC = \angle BOD$

And $\angle AOD = \angle BOC$

$\Rightarrow$ Now ray OA stands on line CD

$\therefore \angle AOC + \angle AOD = 180^{\circ}$ … (i) (linear pair angles)

Similarly, can we write

$\angle AOD + \angle BOD = 180^{\circ}$ … (ii) (linear pair angles)

From equation (i) and (ii) comparing

$\angle AOC + \angle AOD = \angle AOD + \angle BOD$

$\Rightarrow \angle AOC = \angle BOD$

Similarly, we can prove that $\angle AOD = \angle BOC$

Hence Proved.

Question:2

Bisectors of interior $\angle B$ and exterior $\angle ACD$ of a $\triangle ABC$ intersect at the point T. Prove that $\angle BTC = \frac{1}{2} \angle BAC$ .

Answer:

According to the question,
Bisectors of interior $\angle B$ and exterior $\angle ACD$ of a $\triangle ABC$ intersect at the point T
$\angle TBC =\frac{1}{2} \angle ABC$ … (1)
And $\angle TCD =\frac{1}{2} \angle ACD$ … (2)

Now from $\triangle ABC$ we have
$\angle BAC + \angle ABC = \angle ACD$ … (3) (exterior angle is equal to the sum of interior opposite angles)
And from $\triangle TBC$ we have
$\angle BTC + \angle TBC = \angle TCD$ (exterior angle is equal to the sum of interior opposite angles)
Or $\angle BTC + \frac{1}{2} \angle ABC =\frac{1}{2} \angle ACD$ using (1 and 2)
Or $\angle BTC = \frac{1}{2}(\angle ACD - \angle ABC)$
Using (3), $\angle ACD - \angle ABC = \angle BAC$
So,

$\angle BTC = \frac{1}{2} \angle BAC$

Hence Proved

Question:3

A transversal intersects two parallel lines. Prove that the bisectors of any pair of corresponding angles so formed are parallel.

Answer:

Given: A transversal EF cuts two parallel line AB and CD at point G & H. GL and HM are bisectors of angles

To prove: $GL \parallel HM$
Proof: $\angle EGB = \angle GHD$ (Corresponding angles)
$\frac{1}{2} \angle EGB =\frac{1}{2} \angle GHD$

$\angle EGL = \angle GHM$
These are the corresponding angle formed by the line GL and HM, where EF is the transversal.

$\therefore GL \parallel HM$

Hence proved

Question:4

Prove that through a given point, we can draw only one perpendicular to a given line. [Hint: Use proof by contradiction].

Answer:

Given: Consider a line R and a point P

Construction:
Draw two lines (m and n) passing through P which are perpendicular to line R.
To prove: Only one perpendicular line can be drawn through a point P
Proof: In $\triangle APB$
$\angle A + \angle P + \angle B = 180^{\circ}$ {angle sum property}

$90^{\circ} + \angle P + 90^{\circ} = 180^{\circ}$
$\angle P = 180 - 180^{\circ}$
$\angle P = 0^{\circ}$
So lines n and m will coincide
Therefore we can draw only one perpendicular to a given line.

Hence proved

Question:5

Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. [Hint: Use proof by contradiction].

Answer:

Given: Let line x and y are two intersecting lines. Let n and p be another two lines which are perpendicular to x and y

To prove: n and p intersect at a point
Proof: Let lines n and p are not intersecting then $n \parallel p$ … (1)
Since n and p are parallel and n is perpendicular to x and p is perpendicular to y respectively

So, $x \parallel y$
But, it is a contradiction as it is given that x and y are two intersecting lines
Thus our assumption is wrong.
n and p intersect at a point

Hence proved

Question:6

Prove that a triangle must have at least two acute angles.

Answer:

It is given that a triangle must have at least two acute angles.
An acute angle is less than 90 degrees
Let us assume that a triangle does not have two acute angles.
So, it has two angles that are either right angles (=90 degrees) or obtuse angles (greater than 90 degrees)
So let two right angles are present,
So using angle sum property of a triangle, third angle must be zero which is not possible.
Also let one angle be right and one be obtuse. We can take the smallest obtuse angle, i.e.,
So using angle sum property of a triangle, third angle must be negative which is not possible.
Again, if both the angles are obtuse the third angle must be negative which is not possible.
So it is necessary for a triangle to have at least two acute angles.

Hence proved

Question:7

In Fig. 6.17, $\angle Q > \angle R$ , PA is the bisector of $\angle QPR$ and $PM \perp QR$ . Prove that $\angle APM = \frac{1}{2} (\angle Q -\angle R)$ .

Answer:

Given : In $\triangle PQR$ , $\angle Q > \angle R$

PA is the bisector of $\angle QPR$ and $PM \perp QR$ .

To prove : $\angle APM =\left ( \frac{1}{2} \right ) (\angle Q -\angle R)$
Proof: Since PA is the bisector of $\angle P$ , we have

$\angle APQ = \left ( \frac{1}{2} \right )\angle P$ … (i)
In right angled $\triangle PMQ$ we have

$\angle Q + \angle MPQ + 90^{\circ} = 180^{\circ}$ (Angle sum property)

$\Rightarrow \angle MPQ = 90^{\circ} - \angle Q$ … (ii)
Now
$\angle APM = \angle APQ - \angle MPQ$
$\angle APM =\frac{1}{2} \angle P - (90 - \angle Q)$ using (i) & (ii)
$\angle APM = \frac{1}{2} \angle P - 90^{\circ} + \angle Q$
$\angle APM = \frac{1}{2}\angle P -\frac{1}{2} (\angle P + \angle R + \angle Q) + \angle Q$ Since $90^{\circ} = \frac{1}{2}(\angle P + \angle R + \angle Q)$
$\angle APM = \frac{1}{2}\angle P -\frac{1}{2} \angle P - \frac{1}{2} \angle R-\frac{1}{2} \angle Q + \angle Q$
$\angle APM =\frac{1}{2} (\angle Q -\angle R)$

Hence proved

Essential Topics of NCERT Exemplar Solutions Class 9 Maths Chapter 6:

The major topics covered through the NCERT exemplar Class 9 Maths solutions chapter 6 are as follows:

◊ Lines are discussed in detail with the understanding of line segments.

◊ Concepts of parallel lines.

◊ Identify if two lines are intersecting or not.

◊ Concepts of collinearity of three or more than three points.

◊ NCERT exemplar Class 9 Maths chapter 6 solutions discusses the angles in detail.

◊ It includes the understanding of right-angle, acute angle, obtuse angle, complementary angles, et cetera.

◊ The chapter includes knowledge of adjacent angles, vertically opposite angles, and their relations.

◊ This chapter talks about the angle sum property of a triangle which says that all three angles sum to 180°.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate geometry

Chapter 4 Linear equations in Two Variable

Chapter 5 Introduction to Euclid’s Geometry

Chapter 7 Triangles

Chapter 8 Quadrilaterals

Chapter 9 Area of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 6:

These Class 9 Maths NCERT exemplar chapter 6 solutions provide a student with basic concepts of Lines and Angles to progress with geometry learning. In this chapter, Lines and Angles are explained separately. Students can easily attempt and solve other books such as RS Aggarwal Class 9 Maths, NCERT Class 9 Maths, RD Sharma Class 9 Maths, etcetera.

NCERT exemplar Class 9 Maths solutions chapter 6 pdf download feature is available for the students to provide them with coherent learning experience while studying NCERT exemplar Class 9 Maths chapter 6.

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Does the parallel line intersect each other?

No, parallel lines can never intersect each other in other words we can say that Parallel line meets or intersect at infinity.

2. Can any triangle have two interior angles equal to right angle?

No, the sum of all three interior angles has to be 180°, and all the angles will be non-zero angles. If two angles are right-angle, then these two corresponding lines will be parallel to each other and never intersect to form a triangle.

3. Is the learning of lines and triangles useful in the preparation of exams like IITJEE or NEET?

Yes, the concepts of lines and angles are beneficial in understanding problems of physics and mathematics. Mathematics is not the part of the syllabus of NEET; however, knowledge of lines and triangle is very much required in physics

4. How many questions can you expect from Lines and Angles in the final examination?

Generally, a total of 3-4 questions appear yearly on the final examination, which includes the MCQs, very short answers, and occasionally a long-answer type question. These NCERT exemplar Class 9 Maths solutions chapter 6 are sufficient to grasp the concepts and practice required to attempt the paper successfully.

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Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

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Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

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NCERT Exemplar Class 9 Maths Solutions Chapter 6 Lines and Angles

Essential Topics of NCERT Exemplar Solutions Class 9 Maths Chapter 6:

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 6:

Check the Solutions of Questions Given in the Book

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

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