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NCERT Exemplar Class 9 Maths Solutions chapter 11 Constructions

NCERT Exemplar Class 9 Maths Solutions chapter 11 Constructions

Edited By Safeer PP | Updated on Aug 31, 2022 03:02 PM IST

NCERT exemplar Class 9 Maths solutions chapter 11 provides the student with hands-on experience on geometry box instruments and drawing some geometrical figure using it. Geometrical construction plays an essential role in engineering design. At Careers 360, highly skilled subject experts have prepared these NCERT exemplar Class 9 Maths chapter 11 solutions to develop an organised learning flow for the students practicing NCERT Class 9 Maths Book. These NCERT exemplar Class 9 Maths chapter 11 solutions develop a better understanding of the concept of construction as they are detailed and expressive. The syllabus put forward by CBSE for Class 9 has been incorporated in the NCERT exemplar Class 9 Maths solutions chapter 11.

This Story also Contains
  1. NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.1
  2. NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.2
  3. NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.3
  4. NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.4
  5. Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 11:
  6. NCERT Class 9 Exemplar Solutions for Other Subjects:
  7. NCERT Class 9 Maths Exemplar Solutions for Other Chapters:
  8. Features of NCERT Exemplar Class 9 Maths Solutions Chapter 11:

Also, read - NCERT Solutions for Class 9 Maths

NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.1

Question:1

With the help of a ruler and a compass, it is not possible to construct an angle of :
(A) 37.5^{\circ} (B) 40^{\circ} (C) 22.5^{\circ} (D) 67.5^{\circ}

Answer:

As we know that with the help of a ruler and a compass we can construct angles which are bisectors of other angles.
37.5^{\circ} is a bisector of 75^{\circ}.
37.5^{\circ} = \frac{1}{2}\times 75^{\circ}
40^{\circ} is a bisector of 80^{\circ}.
40^{\circ} = \frac{1}{2}\times 80^{\circ}
22.5^{\circ} is a bisector of 45^{\circ}
22.5^{\circ} = \frac{1}{2}\times 45^{\circ}
67.5^{\circ} is a bisector of 135^{\circ}
67.5 ^{\circ}=\frac{1}{2}\times 135^{\circ}
75^{\circ}, 45 ^{\circ}and 135^{\circ} can be constructed with the help of a ruler and a compass as they are all multiples of 15 ^{\circ}.
Hence we can’t construct 40^{\circ} by the use of a ruler and compass.
Therefore option (B) is correct.

Question:2

The construction of a triangle ABC, given that BC = 6 cm, \angle B = 45^{\circ} is not possible when difference of AB and AC is equal to:
(A) 6.9 cm (B) 5.2 cm
(C) 5.0 cm (D) 4.0 cm

Answer:

(A) 6.9 cm
Theorem of inequalities says that the sum of two sides of a triangle is greater than the third side.
So, AC + BC > AB
BC > AB – AC
It is given that BC = 6 cm
6 > AB – AC …..(1)
Only one option does not satisfy equation 1, which is option (A) 6.9 cm
Therefore option (A) is correct.

Question:3

The construction of a triangle ABC, given that BC = 3 cm, \angle C=60^{\circ} is possible when difference of AB and AC is equal to :
(A) 3.2 cm (B) 3.1 cm
(C) 3 cm (D) 2.8 cm

Answer:

(D) 2.8 cm
Theorem of inequalities says that the sum of two sides of a triangle is greater than the third side
So, AB < BC + AB
BC > AB – AC
It is given that BC = 3cm
3 > AB – AC …..(1)
Only one option satisfies the equation 1 which is option (D)
3 > 2.8
Therefore option (D) is correct.

NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.2

Question:1

Write True or False. Give reasons for your answer:
An angle of 52.5^{\circ} can be constructed.

Answer:

True
We know that it is possible to construct both 15^{\circ} and 90^{\circ} with the help of ruler and compass.
An angle of 52.5^{\circ} can be constructed by bisecting angle 105^{\circ}
52.5^{\circ}=\frac{1}{2}\times 105^{\circ}
and 105^{\circ} can constructed by bisector of 120^{\circ} and 90^{\circ}.
Hence the given statement is true.

Question:2

Write True or False. Give reasons for your answer:
An angle of 42.5^{\circ} can be constructed.

Answer:

False
An angle of 42.5^{\circ} cannot be constructed with the help of ruler and compass because it is a bisector of 85° and we can’t draw an angle of 85^{\circ} by the use of a ruler and a compass.
42.5^{\circ}=\frac{1}{2}\times 85^{\circ}
Hence, the given statement is False.

Question:3

Write True or False. Give reasons for your answer:
A triangle ABC can be constructed in which AB = 5 cm, \angle A= 45^{\circ} and BC + AC = 5 cm.

Answer:

Answer
False
Given : AB = 5 cm, \angle A= 45^{\circ} and BC + AC = 5 cm.
According to theorem of inequalities, the sum of two sides of a triangle is greater than the third side.
But here, BC + AC = AB
Hence, the type of triangle can’t be constructed.
Therefore the given statement is false.

Question:4

Write True or False. Give reasons for your answer:
A triangle ABC can be constructed in which BC = 6 cm, \angle C=60^{\circ} and AC – AB = 4 cm.

Answer:

True
Given : BC = 6 cm, \angle C=60^{\circ} and AC – AB = 4 cm.
The theorem of inequalities says that the sum of two sides of a triangle is greater than the third side.
i.e., BC + AB > AC
BC > AC – AB
Putting the given values, we have:
6 > 4 (True)
This type of triangle can be constructed.
Therefore the given statement is True.

Question:5

Write True or False. Give reasons for your answer:
A triangle ABC can be constructed in which \angle B = 105^{\circ}, \angle C = 90^{\circ} and AB + BC + AC = 10 cm.

Answer:

False
Given : \angle B = 105^{\circ}, \angle C = 90^{\circ} , AB + BC + AC = 10 cm.
We know that the sum of interior angles of a triangle is 180°.
But here \angle B+\angle C= 105^{\circ}+90^{\circ}=195^{\circ}
The sum of two angles is greater than 195^{\circ} which is not possible.
Hence, this type of triangle can’t be constructed.
Therefore the given statement is False.

Question:6

Write True or False. Give reasons for your answer:
A triangle ABC can be constructed in which \angle B =60^{\circ}, \angle C=45^{\circ} and AB + BC + AC = 12 cm.

Answer:

Given :\angle B =60^{\circ}, \angle C=45^{\circ} , AB + BC + AC = 12 cm.
We know that the sum of interior angles of a triangle is 180^{\circ}
\angle B +\angle C=60^{\circ}+45^{\circ}=105^{\circ}
105^{\circ}<180^{\circ}
Sum of two angles is less than 180^{\circ}
Also, sum of all the sides is given as 12 cm, which does not contradict any property of a triangle.
Hence this type of triangle can be constructed.
Therefore the given statement is True.

NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.3

Question:1

Draw an angle of 110^{\circ} with the help of a protractor and bisect it. Measure each angle.

Answer:


Steps of Construction:
1. Make an angle of 110^{\circ} by using a protractor \angle ABC = 110^{\circ}
2. Take B as the centre and draw an arc which intersects BC and BA at E and D respectively.
3. Take D and E as centres and draw arcs (without changing the span of the compass) such that they intersect each other at point F.
4. Draw a line BG which passes through F.
Angles :
\angle ABC = 110^{\circ}
\angle ABG = 55^{\circ}
\angle CBG = 55^{\circ}

Question:2

Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?

Answer:


Steps of Construction:
1. Draw a line, AB = 4 cm
2. Take A and B as centers and draw arcs which intersect AB at E and H respectively.
3. Take E as center and draw an arc of same radius as arc of A, B, which intersects the previous arc at F. Then, take the same radius and center F and make an arc which intersects at G.
4. Perform step 3 for H, as a center and make I, J which intersects the previous arc.
5. Now take F and G as centers and draw an arc which intersects each other at C.
6. Take I, J as centers and draw an arc which intersects each other at D.
7. Make lines AC, AD which are perpendicular to AB.
These lines are parallel.

Question:3

Draw an angle of 80^{\circ} with the help of a protractor. Then construct angles of
(i) 40^{\circ} (ii)160^{\circ} and (iii) 120^{\circ}.

Answer:


Steps of Construction :
1. Draw an angle \angle BAC=80^{\circ} with protractor
2. Taking A as the centre, draw a semi-circle with any radius which cuts AB at point G
And cuts AC at point F.
3. Taking F and G as centres, draw arcs which intersect each other at D.
4. Draw AE passing through D.
\angle EAB=40^{\circ}
5. Draw an arc from F with a radius equal to LF which cuts the previous arc at point H.
Then taking H as the centre with the same radius, cut an arc at I.
6. Draw lines AK and AJ passing through I and H respectively.
\angle KAB=160^{\circ}
\angle JAB=120^{\circ}

Question:4

Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect the smallest angle and measure each part.

Answer:


Steps of Construction :
1. Draw a line AB = 4.8 cm
2. With A as center, mark an arc of 3.6 cm radius and with B as center, cut the arc with 3 cm radius at point C.
3. Draw the line AC and BC which are of length 3.6 cm and 3 cm respectively.
4. With A as center, draw an arc which cuts AB and AC at point F, E respectively.
5. With E and F as center, draw arcs which intersect at G.
6. Draw line AD which passes through G,
D is mid-point of BC.
A is the smallest angles because the angle in front of the smallest side is the smallest.
\angle CAB =38^{\circ}, \angle ABC = 50^{\circ}, \angle ACB =92^{\circ}

Question:5

Construct a triangle ABC in which BC = 5 cm, \angle B = 60^{\circ} and AC + AB = 7.5 cm.

Answer:


Given : BC = 5 cm, \angle B = 60^{\circ} , AC + AB = 7.5 cm
Steps of Construction:
1. Draw a line BC = 5 cm
2. Draw an angle of 60^{\circ} at B
3. Mark an arc on line of angle 60° with radius 7.5 cm at point D.
4. Join DC
5. Bisect DC which intersects BD at point A
6. Join AC
DABC is the required triangle.

Question:6

Construct a square of side 3 cm.

Answer:

Given : side = 3 cm

Steps of Construction :
We know that in square each angle is of 90^{\circ} and all sides are equal.
1. Draw AB = 3 cm
2. Draw lines AE and BF with 90^{\circ} angles at A and B respectively.
3. With centers A and B, cut line AE and BF with arcs of radius 3 cm at points C and D respectively.
4. Joint CD
ABCD is a square of 3 cm.

Question:7

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.

Answer:

Given : adjacent sides are of lengths 5 cm and 3.5 cm.

Steps of Construction :
  1. Draw line AB = 5 cm
  2. Draw lines AD and BC with 90° angles at A and B respectively.
3. Mark arc of radius 3.5 cm on AD, BC which intersect AD at E and BC at F.
4. Join EF
ABFE is required rectangle.

Question:8

Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45^{\circ}.

Answer:

Given : Side = 3.4 cm, Angle = 45^{\circ}

Steps of Construction :

  1. Draw a line AB = 3.4 cm
  2. Draw lines AF and BE with 45^{\circ} angles at A and B respectively.
3. Draw arcs with center A and B of radius 3.4 cm which intersect AF at D and BE at C
4. Join DC
ABCD is a rhombus of side 3.4 cm and of angle 45^{\circ}


NCERT Exemplar Class 9 Maths Solutions Chapter 11-Exercise 11.4

Question:1

Construct the following and give justification :
A triangle if its perimeter is 10.4 cm and two angles are 45^{\circ} and 120^{\circ}.

Answer:

Steps of Construction :
  1. Draw a line PQ = 10.4 cm
  2. Draw line PR with 45^{\circ} angle at P.
\angle RPQ=45^{\circ}
  1. Draw line QS with 120^{\circ} angle at Q.
\angle SPQ=120^{\circ}
3. Bisect \angle RPQ and \angle SPQ and then let their bisectors meet at point A
4. Bisect line AQ and AP which meets PQ at C and B respectively.
5. Join A with B and C.
ABC is the required triangle.
Justification:
Since B and C lies on PQ
Hence, PB + BC + CQ = AB + BC + AC = PQ
\angle BAP= \angle APB …..(i) { in DAPB, AB = PB}
\angle ABC= \angle BPA+ \angle APB [\angle ABC is an exterior angle of DAPB]
\angle ABC= 2 \angle APB = \angle RPQ {from (i),\angle RPQ= \angle RPA + \angle APQ}
Also, \angle CAQ = \angle AQC [in DAQC ; AC = CQ]
\angle ACB = \angle CAQ + \angle AQC [\angle ACB is exterior angle of DAQC]
\angle ACB = 2 \angle CAQ = \angle SQP [\AQ is a bisector of \angle SQP]

Question:2

Construct the following and give justification :
A triangle PQR given that QR = 3cm, \angle PQR=45^{\circ} and QP – PR = 2 cm.

Answer:


Given : QR = 3cm, \angle PQR=45^{\circ} , QP – PR = 2 cm.

Steps of Construction :
1. Draw line QR = 3 cm
2. Draw line QB with 45^{\circ} angle at Q.
3. Make an arc on QB of 2cm radius intersecting at A
4. Join AR
5. Bisect AR and extend it so it intersects QB at P
6. Join PR
PQR is the required triangle.

Justification:
P lies on perpendicular bisector of AB.
Hence, PA = PQ
QA = PQ – PA
QA = PQ – PR \left (\because PA = PR \right )

Question:3

Construct the following and give justification :
A right triangle when one side is 3.5 cm and the sum of other sides and the hypotenuse is 5.5 cm.

Answer:

Given : Side = 3.5 cm, sum of other side and hypotenuse = 5.5 cm

Steps Construction :
  1. Draw line BC = 3.5 cm
  2. Draw line BE with 90^{\circ} angle at B.
3. Cut BE by 5.5 cm radius arc at D.
4. Join DC
5. Bisect DC which intersects BD at point A
6. Join AC
ABC is the required triangle.

Justification :
In DBDC and DABC
AD = AC
Sum of AB + AC = 5.5

Question:4

Construct the following and give justification :
An equilateral triangle if its altitude is 3.2 cm.

Answer:

Given : An equilateral triangle with Altitude : 3.2 cm

Steps of Construction :
1. Draw a line XY and take a point D on it .
2. Draw line DE with 90^{\circ} angle at D.
3. Cut the line DE with arc of radius 3.2 cm at point A
4. Make an angle of 30^{\circ} at A which cuts XY at B.
5. Now find the length of BD and as center D cut the line XY at point C with radius equal to BD.
6. Join AC
ABC is the required triangle.

Justification :
Here \angle A = \angle BAD + \angle CAD
=30^{\circ}+30^{\circ}=60^{\circ}
AD\perp XY
In DABD,\angle BAD + \angle DBA + \angle ADB = 30^{\circ}+60^{\circ}+90^{\circ}=180^{\circ}
\angle A = \angle B = \angle C = 60^{\circ}
Hence it is an equilateral triangle.

Question:5

Construct the following and give justification :
A rhombus whose diagonals are 4 cm and 6 cm in lengths.

Answer:

Given : Rhombus with diagonals length = 4 cm, 6 cm

Steps of Construction :
1. Draw a line AB = 4 cm
2. Bisect the AB using another line EF which intersects it at point X
E X F is perpendicular to AB
3. As center X mark an arc of 3 cm as the radius on XF and XE which intersects at C and D respectively.
4. Join AD, AE, BC, BD
ACBD is the required rhombus.
Justification :
DC\perp AB hence DA = DB, AC = BC …..(1)
{every point on perpendicular bisector of line is equidistance from end points of line}
\angle DXB=90^{\circ}
Also, XD = XC = 3 cm
Thus, AB is perpendicular to DC
BD = BC …..(2)
From 1 and 2, ABCD is a rhombus.


Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 11:

The chapter on Constructions covers the following topics through NCERT exemplar Class 9 Maths solutions chapter 11::

  • Students will learn to draw different angles and triangles with the help of a given set of information.
  • The students will learn to draw angular bisectors of any given angle.
  • In this chapter construction of a triangle with given side lengths or angles is explained.
  • NCERT exemplar Class 9 Maths solutions chapter 11 explains that you can draw any triangle with the three pieces of information. One information must be about the side length of the triangle, and the other two data can be about sides or angles.

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 11:

These Class 9 Maths NCERT exemplar chapter 11 solutions provide a hands-on experience to students with geometrical instruments like scale, compass, protractor, et cetera. Geometrical construction is a compelling mathematical branch. However, it is not directly used in higher-level mathematics or engineering entrance examinations, but its knowledge will be instrumental in engineering design and architecture. Students of Class 9 can use these solutions as reference material to better study construction-based practice problems. These Class 9 Maths NCERT exemplar solutions chapter 11 Construction provides ample questions to understand the concept better and elaborate solutions to clarify the doubts.

NCERT exemplar Class 9 Maths solutions chapter 11 pdf download will be made available soon so that students can use NCERT exemplar Class 9 Maths solutions chapter 11 offline. Also, students can use the page download option.

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)

1. If three angles of a triangle are given, can we draw a unique triangle?

No, we cannot draw a unique triangle by knowing only interior angles. We can draw similar triangles having same interior angles. Therefore, there will be multiple triangles having same interior angles.

2. Can we divide any line segment in three equal parts?

Yes, we can divide any line segment in three equal parts by construction. For this refer NCERT exemplar Class 9 Maths solutions chapter 11 pdf download.

3. Can we divide any angle in four equal parts with the help of construction?

Yes, we can do that. From NCERT exemplar Class 9 Maths solutions chapter 11 pdf download, you can learn to draw angular bisector of any angle. Once you learn to draw angle bisector, then attempt drawing bisectors of these two half angles.

Thus, you will get four equal parts of any angle.

4. Can we construct any triangle of side lengths 2 m, 3 m and 6 m?

No, we cannot construct a triangle with the sides given. We know that some of two sides must be more than the third side in any triangle, which is not true in the given case.

Therefore, it is impossible to draw a triangle with these three sides

5. What type of questions are expected from the chapter on Constructions?

Generally, you can expect 1 question, which can be either long-short answer type or long answer-type question from Constructions in the final examination.  NCERT exemplar Class 9 Maths solutions chapter 11 consists of a step-by-step solution to each construction problem, and a thorough understanding of these solutions can lead to higher marks in the examination.

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