# Constructions Chapter 11 Class 9 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 9 Maths Chapter 11‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 9 Maths Chapter 11 Exercise Solution’ using the download button.

### 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

### 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.