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NCERT Solutions for Class 10 Maths Chapter 11 : Constructions

Construction Class 10 NCERT Solutions for Maths Chapter 11 are made by our team of subject experts at Toppr in a detailed manner. NCERT Solutions for class 10 maths chapter 11 Constructions are curated strictly in accordance with the CBSE Curriculum and the exam pattern. The NCERT textbook questions are answered in a way to provide you with a better understanding of the concepts in a systematic and step-by-step manner. It also contains the appropriate diagrams to explain the concepts to the students. As Class 10 exams are Board exams, NCERT Solutions provided by Toppr are the best study material to excel in the exams. These solutions will not only help the students in preparing for the board exams but also for the Olympiads. Also, the MCQs and long and short questions are all answered according to the weightage and the exam pattern. With the help of NCERT Solutions for Class 10 Maths Chapter 11 Constructions you can also analyze your shortcomings and work on them before the exams. These are the best resources designed after proper research and study to assist the students in scoring good marks.

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Introduction

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Exercise 11.1

Question 1

Draw a line segment of length

7.6

cm and divide it in the ratio

5 : 8

. Measure the two parts.

Medium

Solution

Verified by Toppr

D i v i d e a l i n e s e g m e n t o f

7.6 c m,

l e n g t h i n t h e r a t i o

5 : 8

a n d m e a s u r e

m : n = 5 : 8

m + n = 5 + 8 = 13

A C : C B = 5 : 8

b y m e a s u r e m e n t

: A C = 3 c m, C B = 4.6 c m

S t e p s o f c o n s t r u c t i o n

1 . D r a w a n y r a y

A B

s u c h t h a t

A B = 7.6 c m

2 . D r a w

A

X

r a y a t p o i n t

A

m a k i n g a c u t e a n g l e

3

L o c a t e t h e p o i n t s

A_{1}, A_{2}, A_{3}, . . . . . . . . A_{13}

f r o m

A

4

J o i n

B A_{13}

D r a w

B A_{13} ∣ ∣ A_{5} C

N o w

A C : C B = 5 : 8

I f m e a s u r e d

A C = 3 c m, C B = 4.6 c m

Question 2

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are

\frac{3}{5}

time of the corresponding sides of the given triangle.

Medium

Solution

Verified by Toppr

1) Draw

B C = 4 c m

2) Cut arc of

5

cm from B and arc of

6

cm from C. Mark point of intersection as A.

3) Draw BD making an acute angle with BC. Mark 5 equal arcs on BD.

4) Join

B_{5}

to C, draw a line

B_{3} C^{'}

parallel to

B_{5} C

5) Draw a line

C^{'} A^{'}

parallel to

C A

A^{'} B C^{'}

ia the required triangle.

Question 3

Construct a triangle of sides

5 c m, 6 c m

and

7 c m

, then construct a triangle similar to it, whose sides are

\frac{2}{3}

of corresponding sides of the first triangle.

Medium

Solution

Verified by Toppr

Making the triangle

Draw a line segment AB =

5

Draw an arc at

6

cm from point A.

Draw an arc at

7

cm from point B so that it intersects the previous arc.

Joint the point of intersection from A and B.

This gives the required triangle ABC.

Dividing the base:

Draw a ray AX at an acute angle from AB

Plot three points on AX so that;

A A_{1}

A_{1} A_{2}

A_{2} A_{3}

Join

A_{3}

to B.

Draw a line from point

A_{2}

so that this line is parallel to

A_{3}

B and intersects AB at point B'

Draw a line from point B' parallel to BC so that this line intersects AC at point C'

\Rightarrow △ A B^{'} C^{'}

is the required triangle

Question 4

Draw a line segment PQ= 8 cm. Construct the perpendicular bisector of the line segment PQ. Let the perpendicular bisector drawn meets PQ at point R. Measure the length of PR and QR. Is PR=QR?

Medium

Solution

Verified by Toppr

Steps of Construction :

1 .

With

P

and

Q

as centers, draw arcs on both sides of

P Q

with equal radii. The radius should be more than half the length of

P Q

2 .

Let these arcs cut each other at points

R

and

R S

3 .

Join

R S

which cuts

P Q

D

. Then

R S = P Q

. Also

∠ P O R = 90^{\circ}

Hence, the line segment

R S

is the perpendicular bisector of

P Q

as it bisects

P Q

P

and is also perpendicular to

P Q

. On measuring the lengths of

P R = 4

cm,

Q R = 4

cm Since

P R = Q R

, both are

4

cm each

∴ P R = Q R

Question 5

Draw a triangle ABC with

B C = 6

cm, AB

= 5

cm and

∠

ABC

= 6 0^{o}

. Then construct a triangle whose sides are

\frac{3}{4}

of the corresponding sides of the

Δ

ABC.

Medium

Solution

Verified by Toppr

The

Δ A^{'} B C^{'}

whose sides are

\frac{3}{4}

of the corresponding sides of

Δ A B C

can be drawn as follows:

Step 1: Draw a

Δ A B C

with side

B C = 6 c m, A B = 5 c m, ∠ A B C = 6 0^{\circ}

Step 2: Draw a ray

B X

making an acute angle with

B C

on the opposite side of vertex

A

Step 3: Locate

4

points,

B_{1}, B_{2}, B_{3}, B_{4}

on line segment

B X

Step 4: Join

B_{4} C

and draw a line through

B_{3}

, parallel to

B_{4} C

intersecting

B C

C^{'}

Step 5: Draw a line through

C^{'}

parallel to

A C

intersecting

A B

A^{'}

The triangle

A^{'} B C^{'}

is the required triangle.

Question 6

Draw a triangle

A B C

with side

B C = 7 c m, ∠ B = 4 5^{o}, ∠ A = 10 5^{o} .

Then construct a triangle whose side are

\frac{4}{3}

times the corresponding sides of

△ A B C .

Write the construction.

Medium

Solution

Verified by Toppr

In order to construct

△ A B C

, we follow the following steps :

S T E P I

Draw

B C = 7 c m

S T E P I I

B

construct

∠ C B X = 4 5^{o}

and at

C

construct

∠ B C Y = 18 0^{o} - (4 5^{o} - 10 5^{o}) = 3 0^{o}

Suppose

B X

and

C Y

intersect at

A . △ A B C

so obtained is the given triangle.

To construct a triangle similar to

△ A B C

, we follow the following steps.

S T E P I

Construct an acute angle

∠ C B Z

B

on opposite side of vertex

A

△ A B C

S T E P I I

Mark-off four (greater

4

of and

3

\frac{4}{3}

) points

B_{1}, B_{2}, B_{3}, B_{4}

B Z

such that

B B_{1} = B_{1} B_{2} = B_{2} B_{3} = B_{3} B_{4}

S T E P I I I

Join

B_{3}

(the third point) to

C

and draw a line through

B_{4} C

parallel to

B_{3} C

, intersecting the extended line segment

B C

C

S T E P I V

Draw a line through

C

parallel to

C A

intersecting the extended line segment

B A

A

Triangle

A B C

so obtained is the required triangle such that

\frac{A B}{A B} = \frac{B C}{B C} = \frac{A C}{A C} = \frac{4}{3}

Question 7

Draw a right triangle which the sides (other than hypotenuse) are of lengths

4 c m

and

3 c m

. Then construct another triangle whose sides are

\frac{5}{3}

times the corresponding sides of the given triangle.

Hard

Solution

Verified by Toppr

Construct

△ A B C

, the given right angled triangle, as follows:

Step 1: Draw base

B C

of length

4 c m

Step 2: Draw a right angle at point

B

Step 3: Draw an arc of radius

3 c m

on the right angle drawn from

B

and label that point as

A

Step 4: Join

A - C

△ A B C

is the given triangle.

To construct

△ E B D \sim △ A B C

Step 1: Draw a ray

B X

at an acute angle to line

B C

on the opposite side of

A

Step 2: Mark

5

equidistant points

B_{1}, B_{2}, B_{3}, B_{4}, B_{5}

on ray

B X

Step 3: Join

B_{3}

and

C

Step 4: Draw

B_{5} D

parallel to

B_{3} C

and label the intersection on the extension of

B C

D

Step 5: Draw

D E

parallel to

B A

and label the intersection with the extension of

B A

E

△ E B D

is the required triangle.

Exercise 11.2

Question 1

Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle and measure their lengths.

Medium

Solution

Verified by Toppr

Steps of Construction:
1. A circle with radius

6 c m

is drawn taking

O

as centre
2. Point

P

is marked at

10 c m

away from centre of circle.
3. With the half of compass mark

M

which is the midpoint of

O P

.
4. Draw a circle with centre

M

, taking radius

M O

M P

which intersects the given circle at

Q

and

R

.
5. Now join

P Q

and

P R

. These are the tangents of the circle.

We know that, the tangent to a circle is perpendicular to the radius through the point of contact.

∴

△ O P Q, O Q ⊥ Q P

and in

△ O P R, O R ⊥ P R

Hence, both

△ O P Q

and

△ O P R

are right angle triangles.

Applying Pythagoras theorem to both

△ s

, we get:

O P^{2} = O Q^{2} + P Q^{2}

and

O P^{2} = O R^{2} + P R^{2}

O Q = O R = r a d i u s = 6 c m

and

O P = 10 c m

∴ 1 0^{2} = 6^{2} + P Q^{2}

and

1 0^{2} = 6^{2} + P R^{2}

\Rightarrow P Q^{2} = 100 - 36 = 64

and

P R^{2} = 100 - 36 = 64

∴ P Q = P R = 8 c m

Hence, the length of the tangents to a circle of radius

6 c m

, from a point

10 c m

away from the centre of the circle, is

8 c m

Question 2

Construct a tangent to a circle of radius

4

c m

from a point on the concentric circle of radius

6

c m

and measure its length. Also verify the measurement by actual calculation.

Easy

Solution

Verified by Toppr

(1)

Draw two concentric circle

C_{1}

and

C_{2}

with common center

0 and radius

4 c m

and

6 c m

(2)

Take a point P on the outer circle

C_{2}

and join OP.

(3)

Draw the bisector of OP which bisect OP at M'.

(4)

Taking M' as center and OM' as radius draw a dotted circle which

cut the inner circle

C_{1}

at two point M and P.

(5)

Join PM and PP'. Thus, PM and PP' are required tangent.

On measuring PM and PP'.

P M = P P^{'} = 4.4 c m B y c a l c u l a t i o n : I n Δ O M P, ∠ P M O = 9 0^{0} P M^{2} = O P^{2} - O M^{2} (b y p y t h a g o r a s t h e o r e m) P M^{2} = (6)^{2} - (4)^{2} = 36 - 16 = 20 P M^{2} = 20 c m P M = 20 = 4.4 c m H e n c e, t h e l e n g t h o f t h e tan g e n t i s 4.4 c m

Question 3

Draw a circle of radius

3 c m

. Take two points

P

and

Q

on one of its extended on both sides of its centre, each at a distance of

7 c m

on opposites of its centre. Draw tangents to the circle from each of these two points

P

and

Q

Medium

Solution

Verified by Toppr

Step 1: Draw a circle with centre

O

and radius

3 c m

using a compass.

Step 2: Draw a secant passing through the centre. Mark points

P

and

Q

on opposite sides of the centre at a distance of

7 c m

from

O

Step 3: Place the compass on

P

and draw two arcs on opposite sides of

O P

. Now place the compass on

O

and draw two arcs intersecting the arcs drawn from point

P

Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of

O P

. Mark the mid point of

O P

M_{1}

Step 5: From

M_{1}

draw a circle with radius

= M_{1} P = M_{1} O

Step 6: Mark the intersection points of the circle drawn from

M_{1}

with the circle drawn from

O

A

and

B

Step 7: Join

P - A

and

P - B

Step 8: Repeat steps

3

7

for point

Q

and obtain tangents

Q C

and

Q D

P A, P B, Q C

and

Q D

are the required tangents.

Practice more questions

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NCERT Solutions for Class 10 Maths Chapter 11 : Constructions

NCERT Solutions for Class 10 Maths Chapter 11

Class 10 Chapter 11- Constructions covers the topics such as division of a line segment, constructions of tangents to a circle, line segment bisector, etc. The constructions learned by the students in Class 9 such as drawing the perpendicular bisector of a line segment, bisecting an angle, triangle construction etc. will form the basis of Class 10 Constructions. Construction is a part of Geometry in Mathematics and is one of the scoring chapters. NCERT solutions for class 10 also deal with some of the applied facts such as:

The locus of a point that moves at an identical distance from two points, is normal to the line joining both the points.
Perpendicular or Normal means right angles.
A bisector cuts a line segment in two half.

Key Features of NCERT Solutions for Class 10 Maths Chapter 11 - Constructions

Class 10 NCERT Solutions provide a better understanding of the subject and concepts.
The solutions are provided in a step-by-step manner with appropriate diagrams.
These are curated by the experts after thorough research.
They are the best means to evaluate your preparations and overcome your shortcomings.
The Class 10 NCERT Solutions will help the students in board exams as well as Olympiads.
These are absolutely free to download.

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 12 : Area Related to Circles

Chapter 13 : Surface Areas and Volumes

Chapter 14 : Statistics

Chapter 15 : Probability

Frequently Asked Questions on NCERT Class 10 Maths Chapter 11 : Constructions

Question 1. What is meant by the scale factor in the constructions of triangles?

Answer. With respect to the construction of triangles, the scale factor means the ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle.

Question 2. Write a short note on the Construction of Tangents to a Circle.

Answer. There cannot be a tangent to the circle through a point that lies inside a circle. But, if a point lies on the circle, then only one tangent to the circle is possible at this point. This tangent is perpendicular to the radius of the circle through this point. Thus, if we want to draw a tangent at a point on a circle, we can simply draw the radius through this point and draw a line perpendicular to this radius through this point. We will therefore get the required tangent at the point. However, if a point lies outside the circle, there will be two tangents to the circle from this point.

Question 3. What types of constructions are covered in Chapter 11 – Constructions?

Answer. Chapter 11 – Constructions covers the following constructions:

Dividing a line segment in a given ratio.
Constructing a triangle similar to a given triangle as per a given scale factor which may be less than 1 or greater than 1.
Constructing the pair of tangents from an external point to a circle.

Related Chapters

Chapter 1 : Real Numbers

Chapter 2 : Polynomials

Chapter 3 : Pair of Linear Equations in Two Variables

Chapter 4 : Quadratic Equations

Chapter 5 : Arithmetic Progression

Chapter 6 : Triangles

Chapter 7 : Coordinate Geometry

Chapter 8 : Introduction to Trigonometry

Chapter 9 : Some Applications of Trigonometry

Chapter 10 : Circles

Chapter 12 : Area Related to Circles