**1) To Draw the Bisector of a line segment.**

Example: Draw a line segment 5.8 cm long and draw its perpendicular bisector.

*Construction:*1: Draw a line segment AB=5.8 cm by using graduated ruler.

2: With a centre and radius more than half of AB, draw arcs, one on each side of AB.

3: With B centre and some radius as in step 2, draw arcs cutting the arcs drawn in step-2 at E and F respectively.

4: Draw the line segment with E and F as end Points.

The Line segment EF is the required perpendicular bisector of AB.

**2)Â To draw the bisector of a given angle.**

Example: ConstructÂ anÂ angleÂ of atÂ theÂ initialÂ pointÂ ofÂ a given ray and justify the Construction.

*Construction:*1: Draw a ray OA.

2: With O as centre and any suitable radius draw an arc cutting OA at B.

3: With B as centre and same radius cut the previous drawn arc at C and then with C as centre and same radius cut the arc at D.

4:Â WithÂ CÂ asÂ centreÂ andÂ radiusÂ Â moreÂ Â thanÂ halfÂ CDÂ drawÂ anÂ arc.

5: With D as Centre and same radius draw another arc to cut the previous arc at E.

6:Â JoinÂ OE.Â Then

7: Draw the bisector OF of then

By Construction and OF is the bisector of

Therefore,

**3)Â ConstructÂ anÂ equilateralÂ triangle,Â givenÂ itsÂ sideÂ andÂ justifyÂ theÂ construction.**

Example:Â Draw an equilateral triangle of side 4.6 cm

*Construction:*1: DrawÂ Â BCÂ =Â 4.6Â cm

2: With B and C as centres and Radii equal to BC=4.6 cm, draw two arcs on the same side of BC, intersecting each other at A.

3: JoinÂ ABÂ andÂ AC.

JustificationÂ :Â SinceÂ byÂ constructionÂ :

ABÂ =Â BCÂ =Â CAÂ =Â 4.6Â cm

Therefore is an equilateral triangle.

**4) Construction of a triangle when its Base, Sum of the other two sides and one base angle are given.**

Example: ConstructÂ aÂ triangleÂ ABCÂ inÂ whichÂ Â BCÂ =Â 7Â cm,Â andÂ Â ABÂ +Â ACÂ =Â 13Â cm.

*Construction:*1: DrawÂ aÂ rayÂ BXÂ andÂ cutÂ offÂ aÂ lineÂ segmentÂ BCÂ =Â 7Â cm

2: Construct

3: FromÂ BY,Â cutÂ offÂ BDÂ =Â 13Â cm.

4: JoinÂ CD.

5: Draw the perpendicular bisect of CD, intersecting BA at A.

6: JoinÂ AC.

TheÂ triangleÂ ABCÂ thusÂ obtainedÂ isÂ theÂ requiredÂ triangle.

**5) Construction of a triangle when its base, difference of the other two sides and one base angle are given.**

Example:Â ConstructÂ aÂ triangleÂ PQRÂ inÂ whichÂ Â QRÂ =Â 6Â cm andÂ PRÂ â€“Â PQÂ =Â 2 cm.

*Construction:*1: Draw a QX and Cut off a line segment QR= 6 cm from it.

2: Construct a ray QY making an angle of with QR and Produce YQ to form a line YQY'

3: CutÂ offÂ aÂ lineÂ segmentÂ QSÂ =Â 2cmÂ fromÂ QY'.

4: JoinÂ RS.

5: Draw perpendicular bisector of RS intersecting QY at a ponit P.

6: JoinÂ PR.

ThenÂ PQRÂ isÂ theÂ requiredÂ triangle.

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