CONSTRUCTIONS.

Construction 11.1 : To construct the bisector of a given angle.

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

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 DE, draw arcs to 

 2

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.

Join 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, ∆BEF ≅ ∆BDF (SSS rule)

This gives ∠EBF = ∠ DBF (CPCT)

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.

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.

L e t   u s   s e e   h ow  t h i s  me t h o d   g i v e s   u s   t h e

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, ∆  PAQ  ≅ ∆ 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, ∆ PMA ≅ ∆ 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.

Construction 11.3 : To construct an angle of 60

 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.

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, ∆ EAD is an equilateral  triangle  and the  ∠ EAD,  which is the same as

∠ CAB is equal to 60

Construction 11.4 : To construct a triangle, given its base, a base angle and sum

of other two sides.

Given the base BC, a base angle, say ∠B and the sum AB + AC of the other two sides

of a triangle ABC, you are required to construct it.

Steps of Construction :

1. Draw the base BC and at the point B make an

angle, say XBC equal to the given angle.

2. Cut a line segment BD equal to AB + AC from

the ray BX.

3. Join DC and make an angle DCY equal to ∠BDC.

4. Let CY intersect BX at A (see Fig. 11.4).

Then, ABC is the required triangle.

Let us see how you get the required triangle.

Base BC and ∠B are drawn as given. Next in triangle

ACD,

∠ACD = ∠ ADC (By construction)

Therefore, AC = AD and then

AB = BD – AD = BD – AC

AB + AC = BD

Alternative method :

Follow the first two steps as above. Then draw

perpendicular bisector PQ of CD to intersect BD at

a point A (see Fig 11.5). Join AC. Then ABC is the

required triangle. Note that A lies on the perpendicular

bisector of CD, therefore AD = AC.

Remark : The construction of the triangle is not

possible if the sum AB +  AC ≤ BC

Construction 11.5 : To construct a triangle given its base, a base angle and the

difference of the other two sides.

Given the base BC, a base angle, say  ∠B and the difference of other two sides

AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are

following two cases:

Case (i) : Let AB > AC that is AB – AC is given.

Steps of Construction :

1. Draw the base BC and at point B make an angle

say XBC equal to the given angle.

2. Cut the line segment BD equal to AB – AC from ray BX.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4.  L e t   i t   i n t e r s e c t   BX  a t   a   p o i n t  A.   J o i n  A C

(see Fig. 11.6). then ABC is the required triangle.

Let us now see how you have obtained the required triangle ABC.

Base BC and ∠B are drawn as given. The point A lies on the perpendicular bisector of

DC. Therefore, AD = AC

So, BD = AB – AD = AB – AC.

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

Steps of Construction : 1. Same as in case (i).

2. Cut line segment BD equal to AC – AB from the

line BX extended on opposite side of line segment BC.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4. Let PQ intersect BX at A. Join AC (see Fig. 11.7).

Then, ABC is the required triangle.

You can justify the construction as in case (i).

To construct a triangle, given its perimeter and its two base

Given the base angles, say

the triangle ABC.

∠ B and ∠ C and BC + CA + AB, you have to construct

Steps of Construction :

1. Draw a line segment, say XY equal to BC + CA + AB.

2. Make angles LXY equal to

3. Bisect

[see Fig. 11.8(i)].

∠B and MYX equal to ∠C.∠ LXY and ∠ MYX. Let these bisectors intersect at a point A

Fig. 11.8 (i)

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

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

[see Fig 11.8(ii)].

Fig. 11.8 (ii)

Then ABC is the required triangle. For the justification of the construction, you

observe that, B lies on the perpendicular bisector PQ of AX.

Therefore, XB = AB and similarly, CY = AC.

This gives BC + CA + AB = BC + XB + CY = XY.

Again

∠BAX = ∠AXB (As in Δ AXB, AB = XB) and

∠

Similarly,

ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY∠ACB = ∠MYX as required.

Construction 11.6 :

angles.