In earlier chapters, the diagrams, which were necessary to prove a theorem or solving 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 which contains the following:

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

(ii) A pair of set - squares, one with angles 90°, 60° and 30° and 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 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. These will then be used to construct certain kinds of triangles.

As you have learnt how to construct a circle, the perpendicular bisector of a line segment, angles of `30°, 45°, 60°,` `90°` and `120°`, and the bisector of a given angle, without giving any justification for these constructions.

Here, you will construct some of these, with reasoning behind, why these constructions are validtions

`color{blue} text(Construction 11.1 :) "To construct the bisector of a given angle."`

Given an angle ABC, we want to construct its bisector.

`bb text(Steps of Construction :)`

1. Taking B as centre and any radius, draw an arc to intersect the rays `BA` and `BC`, say at `E` and `D` respectively [see Fig.11.1(i)].

2. Next, taking `D` and `E` as centres and with the radius more than `1/2` `DE`, draw arcs to intersect each other, say at `F`.

3. Draw the ray `BF` [see Fig.11.1(ii)]. This ray `BF` is the required bisector of the angle `ABC`.



Let us see how this method gives us the required angle bisector.


`DF` and `EF`.

In triangles `BEF` and `BDF`,

`BE = BD` (Radii of the same arc)

`EF = DF` (Arcs of equal radii)

`BF = BF` (Common)

Therefore, `Delta BEF ≅ Delta BDF` (SSS rule)

This gives `∠EBF = ∠ DBF (CPCT)`

`color {blue} text(Construction) 11.2 : "To construct the perpendicular bisector of a given line segment."`

Given a line segment AB, we want to construct its perpendicular bisector.

`color {blue} text(Steps of Construction) :`

1. Taking A and B as centres and radius more than `1/2 AB`, draw arcs on both sides of the line segment `AB` (to intersect each other).

2. Let these arcs intersect each other at `P` and `Q`. Join `PQ` (see Fig.11.2).

3. Let `PQ` intersect `AB` at the point `M`. Then line `PMQ` is the required perpendicular bisector of `AB`.

Let us see how this method gives us the perpendicular bisector of `AB`.

Join `A` and `B` to both `P` and `Q` to form `AP, AQ, BP` and `BQ`.

In triangles `PAQ` and `PBQ`,


`AP = BP` (Arcs of equal radii)

`AQ = BQ` (Arcs of equal radii)

`PQ = PQ` (Common)

Therefore, `Delta PAQ ≅ Delta PBQ` (SSS rule)

So, `∠ APM = ∠ BPM (CPCT)`

Now in triangles `PMA` and `PMB`,

`AP = BP` (As before)

`PM = PM` (Common)

`∠ APM = ∠ BPM` (Proved above)

Therefore, `Delta PMA ≅ Delta PMB` (SAS rule)

So, `AM = BM` and `∠ PMA = ∠ PMB (CPCT)`

As `∠ PMA + ∠ PMB = 180°` (Linear pair axiom),

we get

`∠ PMA = ∠ PMB = 90°.`

Therefore, `PM`, that is, `PMQ` is the perpendicular bisector of `AB.`

`color{blue} text(Construction 11.3 :)"To construct an angle of 600 at the initial point of a given ray."`

Let us take a ray `AB` with initial point A [see Fig. 11.3(i)]. We want to construct a ray `AC` such that `∠ CAB = 60°`. One way of doing so is given below.

`color{blue}bb text( Steps of Construction :)`

1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.

2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.

3. Draw the ray AC passing through E [see Fig 11.3 (ii)]. Then `∠ CAB` is the required angle of `60°.` Now, let us see how this method gives us the required angle of `60°.`

Join `DE`.

Then, `AE = AD = DE` (By construction)

Therefore, `Delta EAD` is an equilateral triangle and the `∠ EAD`, which is the same as `∠ CAB` is equal to `60°`.