# Chapter Notes: Constructions CBSE Class 9

## Some Important Points

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 ${45^ \circ }$ 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 $\angle AOE = {90^ \circ }$
7: Draw the bisector OF of $\angle AOE$ then $\angle AOF = {45^ \circ }$
By Construction $\angle AOE = {90^ \circ }$ and OF is the bisector of $\angle AOE$
Therefore, $\angle AOF = {1 \over 2}\angle AOE = {1 \over 2} \times {90^ \circ } = {45^ \circ }$

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 $\Delta ABC$ 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  $\angle B = {75^ \circ }$ 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 $\angle XBY = {75^ \circ }$
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 $\angle Q = {60^ \circ }$ 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 ${60^\circ}$ 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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