# 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) To draw the bisector of a given angle.

Example: Construct an angle of ${45^ \circ }$ at the initial point of 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: With C as centre and radius  more  than half CD draw an arc.
5: With D as Centre and same radius draw another arc to cut the previous arc at E.
6: Join OE. 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) Construct an equilateral triangle, given its side and justify the construction.

Example: Draw an equilateral triangle of side 4.6 cm
Construction: 1: Draw  BC = 4.6 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: Join AB and AC.
Justification : Since by construction :
AB = BC = CA = 4.6 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: Construct a triangle ABC in which  BC = 7 cm,  $\angle B = {75^ \circ }$ and  AB + AC = 13 cm.
Construction: 1: Draw a ray BX and cut off a line segment BC = 7 cm
2: Construct $\angle XBY = {75^ \circ }$
3: From BY, cut off BD = 13 cm.
4: Join CD.
5: Draw the perpendicular bisect of CD, intersecting BA at A.
6: Join AC.
The triangle ABC thus obtained is the required triangle.

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

Example: Construct a triangle PQR in which  QR = 6 cm $\angle Q = {60^ \circ }$ and PR – PQ = 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: Cut off a line segment QS = 2cm from QY'.
4: Join RS.
5: Draw perpendicular bisector of RS intersecting QY at a ponit P.
6: Join PR.
Then PQR is the required triangle.

• Anonymous

Message *I really like this website . It is really helpful to me and for others.

• mohnish

I love this website. The notes are really helpful and understandable, even the complete set of questions are useful.

• I found this site amazing amazing!!!!!!!!!!!!!

It is the best than others

• Anonymous

I found this site amazing amazing!!!!!!!!!!!!!

It is the best than others

• shubham kumar

Message *nice thanx

 Get Full Academic Year Course at Flat Rs 5999/- EnrollNOW