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# Revision Notes on Constructions

## Introduction to Constructions

Constructions tell us about the steps of drawing of perfect geometrical figures like triangle, circle, polygons etc by using geometrical tools with the given measurements.

## Geometry Box

To draw the geometrical figures we need some geometrical instruments which we can find in the geometry box. Some instruments are:

### 1. Graduated scale

This is the scale to make the straight lines. Its one side is marked with centimetres and millimetres and the other side is marked with inches. ### 2. A pair of set squares

It is a set of two squares. One with 90°, 60° and 30° angles and other with angles 90°, 45° and 45°. ### 3. Divider

It helps in measuring the length. ### 4. Compass

It is used to draw the circles and angles. ### 5. Protector

It is used to mark and measure the angles. ## Geometrical Construction

In the Geometrical construction, we use only two instruments – A non graduated ruler also called a Straight Edge and a compass for drawing a geometrical figure. For measurements, we may use a graduated scale and protractor also.

### Basic Constructions

Construction 1: How to construct the bisector of an angle?

If we have to bisect the ∠BOA, then we need to follow these steps to construct the bisector of the angle.

Step 1: Take O as the centre and draw an arc by any radius intersecting the rays OA and OB at X and Y respectively.

Step 2: Now take X and Y as the centre and Draw arcs which intersects each other at a point C with radius more than (1/2) XY. Step 3: Join OC to draw a ray which is the required bisector of the ∠BOA.

In the above example, ∠BOA = 80° and OC bisects it in ∠BOC and ∠COA which is 40° each.

Construction 2: How to construct the perpendicular bisector of a given line segment?

We have to bisect the given line segment AB.

Step 1: Take A and B as the centres and radius more than 1/2 of Ab and draw the arcs on both sides. Step 2: These arcs should intersect each other at C and D. And join CD.

Step 3: Here CD intersects AB at point M so that M is the midpoint of AB and CMD is the required perpendicular bisector of AB.

Join A and B to both C and D to form AC, AD, BC and BD.

In triangles CAD and CBD,

AC = BC (Arcs of equal radii)

AD = BD (Arcs of equal radii)

CD = CD (Common)

Therefore, ∆ CAD ≅ ∆CBD (SSS rule)

So, ∠ ACM = ∠ BCM (CPCT)

Now in triangles CMA and CMB,

AC = BC (As before)

CM = CM (Common)

∠ ACM = ∠ BCM (Proved above)

Therefore, ∆ CMA ≅ ∆ CMB (SAS rule)

So, AM = BM and ∠ CMA = ∠ CMB (CPCT)

As ∠ CMA + ∠ CMB = 180° (Linear pair axiom),

Now we get

∠ CMA = ∠CMB = 90°.

Therefore, CMD is the perpendicular bisector of AB.

Construction 3: How to construct an angle of 60° at the initial point of a given ray?

We have to draw an angle of 60° at the given point P.

Step 1: Take P as the centre and draw an arc of any radius which intersects PQ at point B. Step 2: Now Take B as a centre and draw an arc with the same radius as before which intersects the previous arc at point A.

Step 3: Now draw a ray PR which passé through Point A and the ∠RPQ is the required angle of 60°.

Join AB.

Then, AP = AB = PB (By construction)

Therefore, ∆ ABP is an equilateral triangle and the ∠ APB, which is the same as ∠ RPQ is equal to 60°.

## Construction of Triangles

By using the properties of the triangle and the above basic constructions we can construct triangles.

Construction 1: How to construct a triangle if its base, a base angle and sum of other two sides are given?

Given the base BC, a base angle ∠B and the sum of other two sides AB + AC of a triangle ABC, now we need to construct it. Steps of Construction:

Step 1: First of all, draw the base BC and at the point B make an∠XBC equal to the given angle.

Step 2: From the ray BX, cut the line segment BD = AB + AC.

Step 3: Join DC now which make ∠DCY = ∠BDC.

Step 4: When CY intersect BX at A then it form the required triangle i.e. ABC.

Construction 2: How to construct a triangle if its base, a base angle and the difference of the other two sides are given?

Given the base BC, a base angle i.e. ∠B and the difference of the other two sides AB – AC or AC – AB, we need to construct the triangle ABC. There could be two cases:

Case (i): If AB > AC that is AB – AC is given. Steps of Construction:

Step 1: Draw the base BC which is equal to a cm and at point B make an∠XBC = x°.

Step 2: From ray, BX cut the line segment BD which is equal to AB – AC.

Step 3: Join DC and draw the perpendicular bisector of DC.

Step 4: This perpendicular bisector intersects BX at a point A. By joining A to C we get the required triangle i.e. ∆ABC.

Case (ii): If AB < AC that is AC – AB is given.

Steps of Construction: Step 1: Draw the base BC and at point B make an ∠XBC.

Step 2: From ray, BX cut the line segment BD which is equal to AC – AB from the line BX by extending it on opposite side of line segment BC.

Step 3: Join DC and draw the perpendicular bisector of DC.

Step 4: Let PQ intersect BX at A and by joining A to C, we get the required triangle ∆ABC.

Construction 3: How to construct a triangle if its perimeter and two base angles are given?

Given the base angles, say ∠ B and ∠ C and BC + CA + AB, you have to construct the triangle ABC.

Steps of Construction:

Step 1: Draw a line segment XY = BC + CA + AB.

Step 2: Make ∠LXY = ∠B and ∠MYX = ∠C.

Step 3: Now bisect ∠ LXY and ∠ MYX. These bisectors will intersect at a point A.

Step 4: Draw perpendicular bisectors PQ of AX and RS of AY.

Step 5: Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC.

Then ABC is the required triangle. ### Course Features

• Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution