**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.