NCERT Exemplar Class 9 Maths Solutions chapter 11 Constructions

Constructions involve drawing geometric shapes accurately using only a compass and straightedge. In this chapter, you will learn how to construct angles, bisectors, and triangles using specific rules and steps. For example, to divide a line segment or draw a perpendicular bisector, we use a compass to draw arcs from both ends of the line segment and on both sides. After this, the interesting arcs are joined with the help of a scale. This line is the perpendicular bisector of the given line segment. The NCERT Exemplar Class 9 Chapter 11, Construction contains plenty of questions and solutions to practice. NCERT Exemplar Class 9 Maths Solutions Chapter 11 provides the student with hands-on experience with geometry box instruments and drawing some geometrical figures using them.

At Careers 360, highly skilled subject experts have prepared these NCERT Exemplar Class 9 Maths chapter 11 solutions to develop an organized learning flow for the students practicing the NCERT Class 9 Maths Book. These NCERT Exemplar Class 9 Maths chapter 11 solutions create 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. You can find the syllabus, books, notes, and class-wise solutions here.

NCERT Exemplar Class 9 Maths Solutions Chapter 11: Exercise 11.1 Page: 110, Total Questions: 8

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 $\mathrm{\angle }ABC={110}^{\circ }$
2. Take B as the center and draw an arc that intersects BC and BA at E and D respectively.
3. Take D and E as centers 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 :
$\mathrm{\angle }ABC={110}^{\circ }$
$\mathrm{\angle }ABG={55}^{\circ }$
$\mathrm{\angle }CBG={55}^{\circ }$

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 that intersect AB at E and H respectively.
3. Take E as a center and draw an arc of the same radius as the arc of A, and B, which intersects the previous arc at F. Then, take the same radius and center F and make an arc that 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 that intersects each other at C.
6. Take I, and J as centers and draw an arc that intersects each other at D.
7. Make lines AC and 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 $\mathrm{\angle }BAC={80}^{\circ }$ with protractor
2. Taking A as the center, 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 centers, draw arcs that intersect each other at D.
4. Draw AE passing through D.
$\mathrm{\angle }EAB={40}^{\circ }$
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.
$\mathrm{\angle }KAB={160}^{\circ }$
$\mathrm{\angle }JAB={120}^{\circ }$

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

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

Answer:

Given : BC = 5 cm, $\mathrm{\angle }B={60}^{\circ }$ , 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 the line of angle 60° with a radius of 7.5 cm at point D.
4. Join DC
5. the Bisect DC which iinterestscts 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 lines 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:

Give: 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 intersects AD at E and BC at F.
4. Join EF
ABFE requires a rectangle.

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

Answer:

Given : QR = 3cm, $\mathrm{\angle }PQR={45}^{\circ }$ , 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 $(\because PA=PR)$

Question:3 Construct the following and give justification :
A right triangle is when one side is 3.5 cm and the sum of the other sides and the hypnosis tenure is 5.5 cm.

Answer:

Given: Side the = 3.5 cm, the sum of the 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 a 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 the 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 a radius equal to BD.
6. Join AC
ABC is the required triangle.

Justification :
Here $\mathrm{\angle }A=\mathrm{\angle }BAD+\mathrm{\angle }CAD$
$={30}^{\circ }+{30}^{\circ }={60}^{\circ }$
$AD\perp XY$
In DABD,$\mathrm{\angle }BAD+\mathrm{\angle }DBA+\mathrm{\angle }ADB={30}^{\circ }+{60}^{\circ }+{90}^{\circ }={180}^{\circ }$
$\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C={60}^{\circ }$
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 length.

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 the perpendicular bisector of the line is equidistance from the endpoints of the line}
$\mathrm{\angle }DXB={90}^{\circ }$
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 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 piece of information must be about the side length of the triangle, and the other two pieces of data can be about sides or angles.

NCERT Exemplar Class 9 Maths Solutions Chapter Wise

NCERT Class 9 Exemplar Solutions Subject-Wise:

Given below are the subject-wise exemplar solutions of class 9 NCERT:

• NCERT Exemplar Class 9 Maths Solutions
• NCERT Exemplar Class 9 Science solutions

NCERT Solutions for Class 9 Mathematics: Chapter-wise

NCERT Class 9 Exemplar Solutions for Other Subjects

• NCERT Exemplar Class 9 Maths solutions

• NCERT Exemplar Class 9 Science solutions

NCERT Solution Subject Wise

Here are the subject-wise links for the NCERT solutions of class 9:

• NCERT Solution for Class 9 Science
• NCERT Solution for Class 9 Maths

NCERT Notes Subject Wise

Given below are the subject-wise NCERT Notes of class 9:

• NCERT Notes for Class 9 Science
• NCERT Notes for Class 9 Maths

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 9:

• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9