**NCERT Solutions for Class 9 Maths Chapter 11** Constructions is provided here for the benefit of students of Class 9. Geometry is a fundamental concept that is useful in many fields. Therefore, it is necessary to learn the concept and understand its applications. The best way to learn this concept is by referring to the NCERT Solutions for Class 9 Maths Chapter 11 Constructions. These solutions are designed by teachers who can articulate concepts in an efficient manner to help students ace their second term CBSE Class 9 Maths examination.

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**List of Exercises in class 9 Maths Chapter 11:**

### Access Answers of NCERT Class 9 Maths Chapter 11 â€“ Constructions

Exercise 11.1 Page: 191

**1. Construct an angle of 90Â° at the initial point of a given ray and justify the construction. **

Construction Procedure:

To construct an angle 90Â°, follow the given steps:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

3. With B as a centre with the same radius, mark a point C on the arc DCB.

4. With C as a centre and the same radius, mark a point D on the arc DCB.

5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

6. Finally, the ray OP is joined which makes an angle 90Â° with OP is formed.

Justification

To prove âˆ POA = 90Â°

In order to prove this, draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that

OB = BC = OC

Therefore, OBC is an equilateral triangle

So that, âˆ BOC = 60Â°.

Similarly,

OD = DC = OC

Therefore, DOC is an equilateral triangle

So that, âˆ DOC = 60Â°.

From SSS triangle congruence rule

â–³OBC â‰… OCD

So, âˆ BOC = âˆ DOC [By C.P.C.T]

Therefore, âˆ COP = Â½ âˆ DOC = Â½ (60Â°).

âˆ COP = 30Â°

To find the âˆ POA = 90Â°:

âˆ POA = âˆ BOC+âˆ COP

âˆ POA = 60Â°+30Â°

âˆ POA = 90Â°

Hence, justified.

**2. Construct an angle of 45Â° at the initial point of a given ray and justify the construction.**

Construction Procedure:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

3. With B as a centre with the same radius, mark a point C on the arc DCB.

4. With C as a centre and the same radius, mark a point D on the arc DCB.

5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

6. Finally, the ray OP is joined which makes an angle 90Â° with OP is formed.

7. Take B and Q as centre draw the perpendicular bisector which intersects at the point R

8. Draw a line that joins the point O and R

9. So, the angle formed âˆ ROA = 45Â°

Justification

From the construction,

âˆ POA = 90Â°

From the perpendicular bisector from the point B and Q, which divides the âˆ POA into two halves. So it becomes

âˆ ROA = Â½ âˆ POA

âˆ ROA = (Â½)Ã—90Â° = 45Â°

Hence, justified

Solution:

**(i) 30Â°**

Construction Procedure:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc BC which cuts OA at B.

3. With B and C as centres, draw two arcs which intersect each other at the point E and the perpendicular bisector is drawn.

4. Thus, âˆ EOA is the required angle making 30Â° with OA.

Construction Procedure:

1. Draw an angle âˆ POA = 90Â°

2. Take O as a centre with any radius, draw an arc BC which cuts OA at B and OP at Q

3. Now, draw the bisector from the point B and Q where it intersects at the point R such that it makes an angle âˆ ROA = 45Â°.

4. Again, âˆ ROA is bisected such that âˆ TOA is formed which makes an angle of 22.5Â° with OA

**(iii) 15Â°**

Construction Procedure:

1. An angle âˆ DOA = 60Â° is drawn.

2. Take O as centre with any radius, draw an arc BC which cuts OA at B and OD at C

3. Now, draw the bisector from the point B and C where it intersects at the point E such that it makes an angle âˆ EOA = 30Â°.

4. Again, âˆ EOA is bisected such that âˆ FOA is formed which makes an angle of 15Â° with OA.

5. Thus, âˆ FOA is the required angle making 15Â° with OA.

** 4. Construct the following angles and verify by measuring them by a protractor: **

**(i) 75Â° (ii) 105Â° (iii) 135Â° **

Solution:

**(i) 75Â° **

Construction Procedure:

1. A ray OA is drawn.

2. With O as centre draw an arc of any radius and intersect at the point B on the ray OA.

3. With B as centre draw an arc C and C as centre draw an arc D.

4. With D and C as centre draw an arc, that intersect at the point P.

5. Join the points O and P

6. The point that arc intersect the ray OP is taken as Q.

7. With Q and C as centre draw an arc, that intersect at the point R.

8. Join the points O and R

9. Thus, âˆ AOE is the required angle making 75Â° with OA.

**(ii) 105Â° **

Construction Procedure:

1. A ray OA is drawn.

2. With O as centre draw an arc of any radius and intersect at the point B on the ray OA.

3. With B as centre draw an arc C and C as centre draw an arc D.

4. With D and C as centre draw an arc, that intersect at the point P.

5. Join the points O and P

6. The point that arc intersect the ray OP is taken as Q.

7. With Q and Q as centre draw an arc, that intersect at the point R.

8. Join the points O and R

9. Thus, âˆ AOR is the required angle making 105Â° with OA.

**(iii) 135Â°**

Construction Procedure:

1. Draw a line AOA**â€˜**

2. Draw an arc of any radius that cuts the line AOA**â€˜**at the point B and B**â€˜**

3. With B as centre, draw an arc of same radius at the point C.

4. With C as centre, draw an arc of same radius at the point D

5. With D and C as centre, draw an arc that intersect at the point O

6. Join OP

7. The point that arc intersect the ray OP is taken as Q and it forms an angle 90Â°

8. With B**â€˜ **and Q as centre, draw an arc that intersects at the point R

9. Thus, âˆ AOR is the required angle making 135Â° with OA.

**5. Construct an equilateral triangle, given its side and justify the construction.**

Construction Procedure:

1. Let draw a line segment AB = 4 cm .

2. With A and B as centres, draw two arcs on the line segment AB and note the point as D and E.

3. With D and E as centres, draw the arcs that cuts the previous arc respectively that forms an angle of 60Â° each.

4. Now, draw the lines from A and B that are extended to meet each other at the point C.

5. Therefore, ABC is the required triangle.

Justification:

From construction, it is observed that

AB = 4 cm, âˆ A = 60Â° and âˆ B = 60Â°

We know that, the sum of the interior angles of a triangle is equal to 180Â°

âˆ A+âˆ B+âˆ C = 180Â°

Substitute the values

â‡’Â 60Â°+60Â°+âˆ C = 180Â°

â‡’ 120Â°+âˆ C = 180Â°

â‡’âˆ C = 60Â°

While measuring the sides, we get

BC = CA = 4 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 4 cm

âˆ A = âˆ B = âˆ C = 60Â°

Hence, justified.

Exercise 11.2 Page: 195

**1. Construct a triangle ABC in which BC = 7cm, âˆ B = 75Â° and AB+AC = 13 cm**.

Construction Procedure:

The steps to draw the triangle of given measurement is as follows:

1. Draw a line segment of base BC = 7 cm

2. Measure and draw âˆ B = 75Â° and draw the ray BX

3. Take a compass and measure AB+AC = 13 cm.

4. With B as the centre, draw an arc at the point be D

5. Join DC

6. Now draw the perpendicular bisector of the line DC and the intersection point is taken as A.

7. Now join AC

8. Therefore, ABC is the required triangle.

**2. Construct a triangle ABC in which BC = 8cm, âˆ B = 45Â° and ABâ€“AC = 3.5 cm**.

Construction Procedure:

The steps to draw the triangle of given measurement is as follows:

1. Draw a line segment of base BC = 8 cm

2. Measure and draw âˆ B = 45Â° and draw the ray BX

3. Take a compass and measure AB-AC = 3.5 cm.

4. With B as centre and draw an arc at the point be D on the ray BX

5. Join DC

6. Now draw the perpendicular bisector of the line CD and the intersection point is taken as A.

7. Now join AC

8. Therefore, ABC is the required triangle.

**3. Construct a triangle PQR in which QR = 6cm, âˆ Q = 60Â° and PRâ€“PQ = 2cm. **

Construction Procedure:

The steps to draw the triangle of given measurement is as follows:

1. Draw a line segment of base QR = 6 cm

2. Measure and draw âˆ Q = 60Â° and let the ray be QX

3. Take a compass and measure PRâ€“PQ = 2cm.

4. Since PRâ€“PQ is negative, QD will below the line QR.

5. With Q as centre and draw an arc at the point be D on the ray QX

6. Join DR

7. Now draw the perpendicular bisector of the line DR and the intersection point is taken as P.

8. Now join PR

9. Therefore, PQR is the required triangle.

**4. Construct a triangle XYZ in which âˆ Y = 30Â°, âˆ Z = 90Â° and XY+YZ+ZX = 11 cm.**

Construction Procedure:

The steps to draw the triangle of given measurement is as follows:

1. Draw a line segment AB which is equal to XY+YZ+ZX = 11 cm**.**

2. Make an angle âˆ Y = 30Â° from the point A and the angle beâˆ LAB

3. Make an angle âˆ Z = 90Â° from the point B and the angle be âˆ MAB

4. Bisect âˆ LAB and âˆ MAB at the point X.

5. Now take the perpendicular bisector of the line XA and XB and the intersection point be Y and Z, respectively.

6. Join XY and XZ

7. Therefore, XYZ is the required triangle

**5. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.**

Construction Procedure:

The steps to draw the triangle of given measurement is as follows:

1. Draw a line segment of base BC = 12 cm

2. Measure and draw âˆ B = 90Â° and draw the ray BX

3. Take a compass and measure AB+AC = 18 cm.

4. With B as centre and draw an arc at the point be D on the ray BX

5. Join DC

6. Now draw the perpendicular bisector of the line CD and the intersection point is taken as A.

7. Now join AC

8. Therefore, ABC is the required triangle.

Also AccessÂ |

NCERT Exemplar for class 9 Maths Chapter 11 |

CBSE Notes for class 9 Maths Chapter 11 |

Chapter 11 Constructions belongs to Unit 4: Geometry. This unit carries a total of 28 marks out of 100. Therefore, this chapter should be studied thoroughly as a majority of the questions could appear from this unit. The important topics that are covered under this chapter are:

- Basic Construction
- Construction of triangles

Constructions is an important chapter included in the second term of Class 9 Maths that helps the students to understand how different shapes are made. It also teaches their implications and their academic relevance. Learn how to construct a bisector, a perpendicular bisector and much more. Explore more about Constructions and learn to solve various kinds of problems only on NCERT Solutions For Class 9 Maths. It is also one of the best academic resources to revise for your second term exams.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 11 Constructions

- Formulas are explained
- Comprehensive format and jargon-free language
- Constant addition of new questions with solutions
- Additional key tips and tricks are provided.
- Access other important learning resources like sample papers.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 11

### How NCERT Solutions for Class 9 Maths Chapter 11 helpful for second term exam preparation?

### Which is the best source for Class 9 Maths second term exam preparation?

### What are the main concepts covered in NCERT Solutions for Class 9 Maths Chapter 11?

1. Basic Construction

2. Construction of triangles

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