NCERT Exemplar Class 10 Maths Chapter 10 Constructions are available here for students in pdf format, which can be downloaded easily. The problems and solutions are provided by our subject experts as per CBSE latest syllabus (2020-2021), to prepare students for the board exam and score good marks.

The solved questions in exemplars provided here have been specifically designed to boost student’s knowledge about the topics and at the same time help them find all the important answers for all the chapter questions. Students who solve 10th class Maths exemplars will be able to stay ahead in the class, as it is a great study tool.

## Class 10 Maths NCERT Exemplar For Constructions

These exemplars problems and solutions are designed by our experts with respect to CBSE syllabus, which covers the following topics of Constructions given below:

- Dividing the line segment in a given ratio
- Construction of a similar triangle
- To construct tangents to a circle

When we go through the chapter “constructions” in Class 10 maths textbook, students will basically explore new constructions concepts which they should learn and master. Understanding this chapter is crucial as students will need to develop proper mathematical reasoning to discern why some constructions work and some don’t.

In order to help students get a clear insight into the concepts of constructions and get familiar with some of the methods of solving problems, free NCERT exemplar for Class 10 Maths Chapter 10 is provided here. To help students practice more for the exam, BYJU’S is also providing online reading materials such as notes, exemplar books, NCERT Maths solutions for class 10 and question papers, which they can make the best use of it. Students are also recommended to solve sample papers and previous year question papers to get an idea of the type of questions asked from chapter Constructions.

## Download PDF of NCERT Exemplar Class 10 Maths Chapter 10 Constructions

### Access Answers to NCERT Exemplar Class 10 Maths Chapter 10

## Exercise 10.1 Page No: 114

**Choose the correct answer from the given four options:**

**1. To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is**

**(A) 8 (B) 10 (C) 11 (D) 12**

**Solution:**

**(D) 12**

According to the question,

A line segment AB in the ratio 5:7

So, A:B = 5:7

Now,

Draw a ray AX making an acute angle ∠BAX,

Mark A+B points at equal distance.

So, we have A=5 and B=7

Hence, minimum number of these points = A+B = 5+7 =12

**2. To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that BAX is an acute angle and then points A _{1}, A_{2}, A_{3},…. are located at equal distances on the ray AX and the point B is joined to**

**(A) A _{12} (B) A_{11} (C) A_{10} (D) A_{9}**

**Solution:**

**(B) A _{11} **

According to the question,

A line segment AB in the ratio 4:7

So, A:B = 4:7

Now,

Draw a ray AX making an acute angle BAX

Minimum number of points located at equal distances on the ray,

AX = A+B = 4+7= 11

A_{1}, A_{2}, A_{3}, ………. are located at equal distances on the ray AX.

Point B is joined to the last point is A_{11}.

**3. To divide a line segment AB in the ratio 5 : 6, draw a ray AX such that **∠**BAX is an acute angle, then draw a ray BY parallel to AX and the points A _{1}, A_{2}, A_{3}, … and B_{1}, B_{2}, B_{3}, … are located at equal distances on ray AX and BY, respectively. Then the points joined are**

**(A) A _{5} and B_{6} (B) A_{6} and B_{5} (C) A_{4} and B_{5} (D) A_{5} and B_{4}**

**Solution:**

** (A) A _{5} and B_{6}**

According to the question,

A line segment AB in the ratio 5:7

So, A:B = 5:7

Steps of construction:

1. Draw a ray AX, an acute angle BAX.

2. Draw a ray BY ||AX, angle ABY = angle BAX.

3. Now, locate the points A_{1},A_{2},A_{3},A_{4} and A_{5} on AX and B_{1},B_{2},B_{3},B_{4},B_{5} and B_{6}

(Because A:B = 5:7)

4. Join A_{5}B_{6}.

A_{5}B_{6} intersect AB at a point C.

AC:BC= 5:6

## Exercise 10.2 Page No: 115

**Write True or False and give reasons for your answer in each of the following:**

**1. By geometrical construction, it is possible to divide a line segment in the ratio **√**3:(1/√3)**

**Solution:**

True

Justification:

According to the question,

Ratio= √3 : ( 1/√3)

On simplifying we get,

√3/ (1/√3) = (√3 x √3)/1 = 3:1

Required ratio = 3:1

Hence,

Geometrical construction is possible to divide a line segment in the ratio 3:1.

**2. To construct a triangle similar to a given △ABC with its sides 7/3 of the corresponding sides of △ABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect to BC. The points B1, B2, …., B7 are located at equal distances on BX, B3 is joined to C and then a line segment B6C ‘ is drawn parallel to B3C where C‘ lies on BC produced. Finally, line segment A‘C‘ is drawn parallel to AC.**

**Solution:**

False

Justification:

Let us try to construct the figure as given in the question.

Steps of construction,

1. Draw a line segment BC.

2. With B and C as centres, draw two arcs of suitable radius intersecting each other at A.

3. Join BA and CA and we get the required triangle ∆ABC.

4. Draw a ray BX from B downwards to make an acute angle ∠CBX.

5. Now, mark seven points B_{1}, B_{2}, B_{3} …B_{7} on BX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7}.

6. Join B_{3}C and draw a line B_{7}C’|| B_{3}C from B_{7} such that it intersects the extended line segment BC at C’.

7. Draw C’A’ ||CA in such a way that it intersects the extended line segment BA at A’.

Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of

∆ABC.

According to the question,

We have,

Segment B_{6}C’ || B_{3}C. But it is clear in our construction that it is never possible that segment B_{6}C’||B_{3}C since the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B_{7}C’ is parallel to B_{3}C.

## Exercise 10.3 Page No: 116

**1. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.**

**Solution:**

Steps of construction:

1. Draw a line segment, AB = 7 cm.

2. Draw a ray, AX, making an acute angle down ward with AB.

3. Mark the points A_{1}, A_{2}, A_{3 }… A_{8} on AX.

4. Mark the points such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = ….., A_{7}A_{8}.

5. Join BA_{8}.

6. Draw a line parallel to BA_{8} through the point A_{3}, to meet AB on P.

Hence AP: PB = 3: 5

**2. Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle?**

**Solution:**

Steps of construction:

1. Draw a line segment AB = 5 cm. Construct a right angle SAB at point A.

2. Draw an arc of radius 12 cm with B as its centre to intersect SA at C.

3. Join BC to obtain ABC.

4. Draw a ray AX making an acute angle with AB, opposite to vertex C.

5. Locate 3 points, A_{1}, A_{2}, A_{3 }on line segment AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3}.

6. Join A_{3}B.

7. Draw a line through A_{2} parallel to A_{3}B intersecting AB at B’.

8. Through B’, draw a line parallel to BC intersecting AC at C’.

9. Triangle AB’C’ is the required triangle.

## Exercise 10.4 Page No: 117

**1. Two line segments AB and AC include an angle of 60° where AB = 5 cm and AC = 7 cm. Locate points P and Q on AB and AC, respectively such that AP = ¾ AB and AQ = ¼ AC. Join P and Q and measure the length PQ. **

**Solution:**

Steps of construction:

1. Draw a line segment *AB = *5 cm.

2. Draw ∠BAZ = 60°.

3. With centre A and radius 7 cm, draw an arc cutting the line AZ at C.

4. Draw a ray AX, making an acute ∠BAX.

5. Divide AX into four equal parts, namely AA_{1 }= A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4}._{ }

6. Join A_{4}B.

7. Draw A_{3}P || A_{4}B meeting AB at P.

8. Hence, we obtain, P is the point on AB such that AP = ¾ AB.

9. Next, draw a ray AY, such that it makes an acute ∠CAY.

10. Divide AY into four parts, namely AB_{1} = B_{1}B_{2 }= B_{2}B_{3} = B_{3}B_{4}.

11. Join B_{4}C.

12. Draw B_{1}Q || B_{4}C meeting AC at Q. We get, Q is the point on AC such that AQ = ¼ AC.

13. Join PQ and measure it.

14. *PQ *= 3.25 cm.

**2. Draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm and angle ABC = 60°, divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD ‘ C‘ similar to triangle BDC with scale factor 4/3. Draw the line segment D‘A‘ parallel to DA where A’ lies on extended side BA. Is A’BC’D’ a parallelogram? **

**Solution:**

Steps of constructions:

1. Draw a line AB=3 cm.

2. Draw a ray BY making an acute ∠ABY=60°.

3. With centre B and radius 5 cm, draw an arc cutting the point C on BY.

4. Draw a ray AZ making an acute ∠ZAX’=60°.(BY||AZ, ∴ ∠YBX’=ZAX’=60°)

5. With centre A and radius 5 cm, draw an arc cutting the point D on AZ.

6. Join CD

7. Thus we obtain a parallelogram ABCD.

8. Join BD, the diagonal of parallelogram ABCD.

9. Draw a ray BX downwards making an acute ∠CBX.

10. Locate 4 points B_{1}, B_{2}, B_{3}, B_{4} on BX, such that BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}.

11. Join B_{4}C and from B_{3}C draw a line B4C’||B3C intersecting the extended line segment BC at C’.

12. Draw C’D’|| CD intersecting the extended line segment BD at D’. Then, ∆D’BC’ is the required triangle whose sides are 4/3 of the corresponding sides of ∆DBC.

13. Now draw a line segment D’A’|| DA, where A’ lies on the extended side BA.

14. Finally, we observe that A’BC’D’ is a parallelogram in which A’D’=6.5 cm A’B = 4 cm and ∠A’BD’= 60° divide it into triangles BC’D’ and A’BD’ by the diagonal BD’.

**3. Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.**

**Solution:**

Steps of constructions:

1. Draw a circle with center O and radius 3 cm.

2. Draw another circle with center O and radius 5 cm.

3. Take a point P on the circumference of larger circle and join OP.

4. Draw another circle with diameter OP such that it intersects the smallest circle at A and B.

5. Join A to P and B to P.

Hence AP and BP are the required tangents.

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