Contructions

8/12/2019 Contructions

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CONSTRUCTIONS


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Basic Constructions

Construction 1

Aim: To construct the bisector of a given angle

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. 1[i]2. Next, taking D and E as centres and with the

radius more than DE, draw arcs tointersect each other, say at F.


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3.Draw the ray BF {See Fig.1[ii]}. This ray BF is the required bisector of the angleABC.

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 is congruent to BDF [SSS rule]

This gives angle EBF= angle DBF [Corresponding Parts of Congruent Triangles]


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Figure 1[i

]


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Figure 1[ii]


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Construction 2

Aim: To construct the perpendicular bisector of a givenline segment.

Steps of Construction:1. Taking A and B as centres and radius more than

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. JoinPQ[See Fig. 2].

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


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Let us see how this method gives us theperpendicular bisector of AB.

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

In triangles PAQ and PBQ,

AP= BP [Arcs of equal radii]

AQ= BQ [Arcs of equal radii]

PQ= PQ [Common]

Therefore, PAQ is congruent to PBQ [SSS rule]

So, angle APM= angle BPM [Congruent Parts ofCongruent Triangles]


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Now in triangles PMA and PMB,

AP= BP [As before]

PM= PM [Common]

APM= BPM [Proved above]

Therefore, PMA is congruent to PMB [SAS rule]

So, AM= BM and angle PMA= angle PMB [CorrespondingParts of Congruent Triangles]

As PMA + PMB = 180 degree [Linear pair axiom]

we getangle PMA= angle PMB= 90 degree.

Therefore, PM, that is, PMQ is the perpendicular bisector ofAB.


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Figure 2


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Construction 3

Aim: To construct an angle of 60 degree at theinitial point of a given ray.

Let us take a ray with initial point A [See Fig.3 {i}].We want to construct a ray AC such that CAB= 60degree. One way of doing so is given below.

Steps of Construction:

1. Taking A as centre and some radius, draw anarc of a circle, which intersects AB, say at a pointD.


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

3. Draw the ray AC passing through E [See Fig. 3 {ii}].

Then CAB is the required angle of 60 degree. Now,let us see how this method gives us the requiredangle of 60 degree.

Join DE.

Then, AE= AD= DE [By construction]Therefore, EAD is an equilateral triangle and theangle EAD, which is the same as angle CAB is equalto 60 degree.


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Figure 3[i]


13/27Figure 3[ii]


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Some Constructions of

TrianglesConstruction 4Aim: To construct a triangle, given its base, a base angle and sum ofother 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 XBCequal to the given angle.

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

3. Join DC and draw the perpendicular bisector PQ of CD.4. Let PQ intersect ray BX at A. Join AC.

Then, ABC is the required triangle.


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Proof: Note that A lies on the perpendicularbisector of CD, therefore AD= AC.

Remark:The construction of the triangle is not

possible if the sum AB+ AC < BC.


16/27Figure 4


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Construction 5

Aim:To construct a triangle given its base, a

base angle and the difference of the othertwo sides.

Given the base BC, a base angle, say B andthe difference of other two sides AB- AC or

AC- AB, you have to construct the triangleABC. Clearly there are following two cases:

Case{i}: Let AB> AC that is AB- AC is given.


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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 perpendicularbisector, say PQ of DC.

4. Let it intersect BX at a point A. JoinAC[See Fig.5 {i}]


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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 pointA lies on the perpendicular bisector of DC.Therefore,

AD= AC

So, BD= AB- AD= AB- AC.


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Figure 5[i]


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Case{ii}: Let AB< AC that is AC- AB is given.


1. Same as in case [i].

2. Cut line segment BD equal to AC- AB from the lineBX extended on opposite side of line segment BC.

3. Join DC and draw the perpendicular bisector, sayPQ of DC.

4. Let PQ intersect BX at A. Join AC [See Fig.5 {ii}]Then, ABC is the required triangle.

We can justify the construction as in case {i}.


22/27Figure 5[ii]


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Construction 6

Aim: To construct a triangle, given its perimeter andits two base angles.

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


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

2. Make angles LXY equal to B and MYX equalto C.


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3. Bisect LXY and MYX. Let these bisectorsintersect at a point A [See Fig. 6{i}]

4. Draw perpendicular bisectors PQ of AX and

RS of AY.5. Let PQ intersect XY at B and RS intersectXY at C. Join AB and AC [See Fig.6{ii}]

Then ABC is the required triangle. For thejustification of the construction, we observethat, B lies on the perpendicular bisector PQof AX.


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

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

Again angle BAX= angle AXB [As inAXB, AB=XB] and

angle ABC= angleBAX+angleAXB= 2 angle AXB= angleLXY

Similarly, angleACB= angleMYX as required.


26/27Figure 6[i]


27/27Figure 6[ii]

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