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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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## Access NCERT Solutions for Class 10 Maths Chapter 11 : Constructions

Exercise 11.1
Question 1
Draw a line segment of length cm and divide it in the ratio . Measure the two parts.
Medium
Solution
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.

.
:
,
.
. .
. . .
,
,.
Question 2
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are time of the corresponding sides of the given triangle.
Medium
Solution
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1) Draw
2) Cut arc of cm from B and arc of 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 to C, draw a line parallel to .
5) Draw a line parallel to .
ia the required triangle.
Question 3
Construct a triangle of sides and , then construct a triangle similar to it, whose sides are of corresponding sides of the first triangle.
Medium
Solution
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Making the triangle

Draw a line segment AB = cm

Draw an arc at cm from point A.

Draw an arc at 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; = =

Join to B.

Draw a line from point so that this line is parallel to 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'

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
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Steps of Construction :
With and as centers, draw arcs on both sides of with equal radii. The radius should be more than half the length of
Let these arcs cut each other at points and
Join which cuts at . Then . Also .
Hence, the line segment is the perpendicular bisector of as it bisects at and is also perpendicular to . On measuring the lengths of cm, cm Since , both are cm each
.
Question 5
Draw a triangle ABC with cm, ABcm and ABC . Then construct a triangle whose sides are of the corresponding sides of the ABC.
Medium
Solution
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The whose sides are of the corresponding sides of can be drawn as follows:

Step 1: Draw a with side

Step 2: Draw a ray making an acute angle with on the opposite side of vertex .

Step 3: Locate points, on line segment .

Step 4: Join and draw a line through , parallel to intersecting at .

Step 5: Draw a line through parallel to intersecting at .

The triangle is the required triangle.

Question 6
Draw a triangle with side Then construct a triangle whose side are times the corresponding sides of Write the construction.
Medium
Solution
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In order to construct , we follow the following steps :

Draw .

At construct and at construct

Suppose and intersect at so obtained is the given triangle.

To construct a triangle similar to , we follow the following steps.

Construct an acute angle at on opposite side of vertex of .

Mark-off four (greater of and in ) points on such that
.

Join (the third point) to and draw a line through parallel to , intersecting the extended line segment at .

Draw a line through parallel to intersecting the extended line segment at .

Triangle so obtained is the required triangle such that

Question 7
Draw a right triangle which the sides (other than hypotenuse) are of lengths and . Then construct another triangle whose sides are times the corresponding sides of the given triangle.
Hard
Solution
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Construct , the given right angled triangle, as follows:
Step 1: Draw base of length .
Step 2: Draw a right angle at point .
Step 3: Draw an arc of radius on the right angle drawn from and label that point as
Step 4: Join .
is the given triangle.

To construct :
Step 1: Draw a ray at an acute angle to line on the opposite side of .
Step 2: Mark equidistant points on ray .
Step 3: Join and .
Step 4: Draw parallel to and label the intersection on the extension of as .
Step 5: Draw parallel to and label the intersection with the extension of as

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
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Steps of Construction:
1. A circle with radius is drawn taking as centre
2. Point is marked at away from centre of circle.
3. With the half of compass mark which is the midpoint of .
4. Draw a circle with centre , taking radius or which intersects the given circle at and .
5. Now join and . 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.

In
and in

Hence, both and are right angle triangles.

Applying Pythagoras theorem to both , we get:
and

and

and

and

Hence, the length of the tangents to a circle of radius , from a point away from the centre of the circle, is .
Question 2
Construct a tangent to a circle of radius from a point on the concentric circle of  radius and measure its length. Also verify the measurement by actual calculation.
Easy
Solution
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Draw two concentric circle and with common center
Take a point P on the outer circle and join OP.
Draw the bisector of OP which bisect OP at M'.
Taking M' as center and OM' as radius draw a dotted circle which
cut the inner circle at two point M and P.
Join PM and PP'.  Thus, PM and PP' are required tangent.
On measuring PM and PP'.

Question 3
Draw a circle of radius . Take two points and on one of its extended on both sides of its centre, each at a distance of on opposites of its centre. Draw tangents to the circle from each of these two points and
Medium
Solution
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Step 1: Draw a circle with centre and radius using a compass.
Step 2: Draw a secant passing through the centre. Mark points and on opposite sides of the centre at a distance of from .
Step 3: Place the compass on and draw two arcs on opposite sides of . Now place the compass on and draw two arcs intersecting the arcs drawn from point .
Step 4: Join the intersection points of the arcs to obtain the perpendicular bisector of . Mark the mid point of as
Step 5: From draw a circle with radius
Step 6: Mark the intersection points of the circle drawn from with the circle drawn from as and .
Step 7: Join and
Step 8: Repeat steps to for point and obtain tangents and

and are the required tangents.
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## 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:
1. The locus of a point that moves at an identical distance from two points, is normal to the line joining both the points.
2. Perpendicular or Normal means right angles.
3. 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.
Related Chapters

### 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:

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