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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