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## NCERT Class 9 Maths Textbook Chapter 11 With Answer Book PDF Free Download

### Chapter 11: Constructions

#### 11.1 Introduction

In earlier chapters, the diagrams, which were necessary to prove a theorem or solve exercises were not necessarily precise. They were drawn only to give you a feeling for the situation and as an aid for proper reasoning.

However, sometimes one needs an accurate figure, for example – to draw a map of a building to be constructed, to design

tools, and various parts of a machine, to draw road maps, etc.

To draw such figures some basic geometrical instruments are needed. You must be having a geometry box that contains the following:

(i) A graduated scale, on one side of which centimeters and millimeters are marked off and on the other side inches and their parts are marked off.

(ii) A pair of a set – squares, one with angles 90°, 60°, and 30° and the other with angles 90°, 45°, and 45°.

(iii) A pair of dividers (or a divider) with adjustments.

(iv) A pair of compasses (or a compass) with the provision of fitting a pencil at one end.

(v) A protractor.

Normally, all these instruments are needed in drawing a geometrical figure, such as a triangle, a circle, a quadrilateral, a polygon, etc. with given measurements.

But a geometrical construction is the process of drawing a geometrical figure using only two instruments – an ungraduated ruler, also called a straight edge and a compass.

In construction where measurements are also required, you may use a graduated scale and protractor also. In this chapter, some basic constructions will be considered

Author | NCERT |

Language | English |

No. of Pages | 9 |

PDF Size | 303 KB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### NCERT Solutions Class 9 Maths Chapter 11 Constructions

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

Construction Procedure:

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

1. Draw a ray OA

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

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

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

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

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

Justification

To prove ∠POA = 90°

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, ∠BOC = 60°.

Similarly,

OD = DC = OC

Therefore, DOC is an equilateral triangle

So that, ∠DOC = 60°.

From the SSS triangle congruence rule

△OBC ≅ OCD

So, ∠BOC = ∠DOC [By C.P.C.T]

Therefore, ∠COP = ½ ∠DOC = ½ (60°).

∠COP = 30°

To find the ∠POA = 90°:

∠POA = ∠BOC+∠COP

∠POA = 60°+30°

∠POA = 90°

Hence, justified.

NCERT Class 9 Maths Textbook Chapter 11 Constructions PDF Free Download