# CBSE Class 9 Maths Revision Notes for Geometric Constructions

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**Constructions**

**To Construct an Angle Equal to a Given Angle**

Given: Any ∠POQ and a point A

Required: To construct an angle at A equal to ∠POQ

**Steps of Construction:**

- With O as centre and any (suitable) radius, draw an arc to meet OP at R and OQ at S.
- Through A draw a line AB.
- Taking A as centre and same radius (as in step 1), draw an arc to meet AB at D.
- Measure the segment RS with compasses.
- With D as centre and radius equal to RS, draw an arc to meet the previous arc at E.
- Join AE and produce it to C, then ∠BAC is the required angle equal to ∠POQ.

**To Bisect a Given Angle**

Given: Any ∠POQ

Required: To bisect ∠POQ.

**Steps of Construction:**

- With O as centre and any (suitable) radius, draw an arc to meet OP at R and OQ at S.
- With R as centre and radius more than half of RS, draw an arc. Also, with S as centre and same radius draw another arc to meet the previous arc at T.
- Join OT and produce it, then OT is the required bisector of ∠POQ.

**To Construct some Specific Angles**

**To construct an angle of 60°**

**Steps of Construction:**

- Draw any line OP.
- With O as centre and any suitable radius, draw an arc to meet OP at R.
- With R as centre and same radius (as in step 2), draw an arc to meet the previous arc at S.
- Join OS and produce it to Q, then ∠POQ = 60°.

**To construct an angle of 30°**

**Steps of Construction**

i. Construct ∠POQ = 60°.

ii. Bisect ∠POQ. Let OT be the bisector of ∠POQ , then ∠POT = 30°

**To construct an angle of 120°**

**Steps of Construction:**

- Draw any line OP.
- With O as centre and any suitable radius, draw an arc to meet OP at R.
- With R as centre and same radius (as in step 2), draw an arc to meet the previous arc at T. With T as centre and same radius, draw another arc to cut the first arc at S.
- Join OS and produce it to Q, then ∠POQ = 120°.

**To construct an angle of 90°**

**Steps of Construction:**

- Construct ∠POQ = 60°
- Construct ∠POV = 120°.
- Bisect ∠QOV. Let OU be the bisector of ∠QOV, then ∠POU = 90°.

**To construct an angle of 45°**

**Steps of Construction:**

- Construct ∠AOP = 90°.
- Bisect ∠AOP.

Let OQ be the bisector of ∠AOP, then ∠AOQ = 45°

** To Draw a Perpendicular Bisector of a Line Segment**

Given: Any line segment PQ.

Required: To draw a perpendicular bisector of line segment PQ.**Steps of Construction:**

- With P as centre, take a length greater than half of PQ and draw arcs one on each side of PQ.
- With Q as centre and same radius (as in step 1), draw two arcs on each side of PQ cutting the previous arcs at A and B.
- Join AB to meet PQ at M, then AB bisects PQ at M, and is perpendicular to PQ, Thus, AB is the required perpendicular bisector of PQ.

**Construction of a Triangle, given its Base, sum of the other two sides and one Base Angle**

To construct ΔABC in which base BC, ÐB and sum AC + AB of other two sides are given.

**Steps of construction:**

- Draw the base BC and at the point B, make an angle, say XBC equal to the given angle.
- Cut a line segment BD = AC + AB from the ray BX.
- Join DC and make angle DCY equal to angle BDC.
- Let CY intersect BX at A.
- ABC is the required triangle.

**Alternate Method**

**Steps of construction:**

- Draw the base BC and at the point B, make an angle, say XBC equal to the given angle.
- Cut a line segment BD = AC + AB from the ray BX.
- Draw perpendicular bisector PQ of CD to intersect BD at a point A. Join AC.

ABC is the required triangle.

**Construction of a Triangle, given its Base, difference of the other two sides and one Base Angle**

To construct ΔABC when the base BC, a base angle B and the difference of other two sides AB - AC or AC - AB are given.

**Case 1**: When AB > AC and AB - AC is given

**Steps of construction:**

- Draw the base BC and at point B make an angle say XBC equal to the given angle.
- Cut the line segment BD equal to AB - AC from ray BX.
- Join DC and draw the perpendicular bisector, say PQ of DC. Let it intersect BX at a point A. Join AC

Then, ABC is the required triangle.

**Case 2**: When AB < AC and AC - AB is given

**Steps of Construction:**

- Draw the base BC and at point B make an angle say XBC equal to the given angle.
- Cut a line segment BD equal to AC - AB from the line BX extended on opposite side of line segment BC.
- Join DC and draw the perpendicular bisector, say PQ of DC.
- Let PQ intersect BX at A. Join AC.

Then, ABC is the required triangle.

**Construction of a Triangle of given Perimeter and Base Angles**

To construct a triangle ABC, when its perimeter, AB + BC + CA, and two base angles, ∠B and ∠C, are given.

**Steps of Construction:**

- Draw a line segment, say XY = BC + CA + AB.
- Construct ∠LXY = ∠B and ∠MYX = C.
- Draw the bisectors of ∠LXY and ∠MYX. Let these bisectors intersect at point A
- Draw a perpendicular bisector PQ of AX and RS of AY.
- Let PQ intersect XY at B and RS intersect XY at C.
- Join AB and AC. Then, ABC is the required triangle.

## Maths Chapters for Revision Notes

## CBSE Class 9 Maths Homework Help

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