## Revision Notes of Chapter 11 Construction Class 9th Math

**Topics in the Chapter**

- Basics of Constructions
- Construction of Angle Bisector
- Construction of Important Angles with some measurement.
- Construction of Perpendicular Bisector
- Construction of a triangle, given its base, difference of the other two sides and one base angle.
- Construction of a triangle of given perimeter and base angles.

**Basics of Constructions**

• A geometrical construction means to draw geometrical figures, such as an angle, a circle, a triangle, a quadrilateral, and a polygon, etc.

• We normally use all or some of the following instruments for drawing geometrical figures:

(i) A protractor

(ii) A pair of compasses

(iii) A pair of set squares

(iv) A pair of dividers

(v) A graduated scale.

**Construction of Angle Bisector**

**1. Steps for construction of Angle Bisector for a given angle.**

__Given:__∠POQ.

__Construction:__To construct the bisector of ∠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 any suitable radius (not necessarily) equal to radius of step 1 (but > 1/2 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.

**2. Construction of Important Angles with some measurement.**

**(i) 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°.

**(ii) To construct an angle of 30°**

__Steps of Construction:__

- Construct ∠POQ = 60° (as above).
- Bisect ∠POQ (as in construction I). Let OT be the bisector of ∠POQ, then ∠POT = 30°.

**(iii) To construct an angle of 120°**

__Steps of Construction:__

- Draw any line segment 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°

**(iv) 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°.

**(v) To construct an angle of 45°**

__Steps of Construction:__

- Construct ∠AOP = 90°
- Bisect ∠AOP.
- Let OQ be the bisector of ∠AOP, then ∠AOQ = 45°.

**3. Construction of Perpendicular Bisector**

__Given:__Any line segment AB.

__Construction:__To construct a perpendicular bisector of line segment AB.

__Steps of Construction:__

- Draw a line segment AB.
- Taking A and B as the centres and radius of more than half the length of AB, draw arcs on both sides of AB.
- Let these arcs intersect each other at points M and N. 4. Join the points of intersection M and N. Thus, MN is the required perpendicular bisector of AB.

**4. Construction of a triangle, given its base, difference of the other two sides and one base angle.**

__Construction:__Construct a triangle with base of length 7·5 cm, the difference of the other two sides is 2·5 cm, and one base angle of 45°.

__Given:__In DABC, base BC = 7·5 cm, the difference of the other two sides, AB – AC or AC – AB = 2·5 cm and one base angle is 45°.

__Required:__To construct the DABC.

**Case (i)**AB – AC = 2·5 cm.

__Steps of Construction:__

- Draw BC = 7·5 cm.
- At B, construct ∠CBX = 45°.
- From BX, cut off BD = 2·5 cm.
- Join CD.
- Draw the perpendicular bisector RS of CD intersecting BX at a point A.
- Join AC. Then DABC is the required triangle.

**Case (ii)**AC – AB = 2·5 cm.

__Steps of Construction:__

- Draw BC = 7·5 cm.
- At B, construct ∠CBX = 45° and produce XB to form a line XBX’.
- From BX’, cut off BD’ = 2·5 cm.
- Join CD’.
- Draw the perpendicular bisector RS of CD’ intersecting BX at a point A.
- Join AC. Then DABC is the required triangle.

**5. Construction of a triangle of given perimeter and base angles.**

__Construction:__Construct a triangle with perimeter 11·8 cm and base angles 60° and 45°.

__Given:__In DABC, AB + BC + CA = 11·8 cm, ∠B = 60° and ∠C = 45°.

__Construction:__To construct the DABC.

__Steps of Construction:__

- Draw DE = 11·8 cm.
- At D, construct ∠EDP = 1/2 of 60° = 30° and at E, construct ∠DEQ = 1/2 of 45° = 22½°.
- Let DP and EQ meet at A. 4. Draw a perpendicular bisector of AD to meet DE at B.
- Draw a perpendicular bisector of AE to meet DE at C.
- Join AB and AC. Then DABC is the required triangle.