## Revision Notes ofÂ 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**

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

â€¢ 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:__âˆ POQ.Â

__Construction:__To construct the bisector of âˆ POQ.Â

__Steps of Construction:__Â Â

- With O as centre and any suitable radius, drawÂ an arc to meet OP at R and OQ at S.
- With R as centre and any suitable radius (notÂ necessarily) equal to radius of step 1 (but > 1/2 RS),Â draw an arc. Also, with S as centre and sameÂ radius, draw another arc to meet the previous arcÂ at T.
- Join OT and produce it, then OT is the requiredÂ bisector of âˆ POQ.Â Â

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

**(i) To construct an angle of 60Â°**

__Steps of Construction:__

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

**(ii) To construct an angle of 30Â°**

__Steps of Construction:__

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

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

__Steps of Construction:__

- Draw any line segment OP.
- With O as centre and any suitable radius, drawÂ 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 asÂ centre and same radius, draw another arc to cutÂ the first arc at S.
- Join OS and produce it to Q, then âˆ POQ = 120Â°

**(iv) To construct an angle of 90Â°**

__Steps of Construction:Â__

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

**(v) To construct an angle of 45Â°**

__Steps of Construction:__

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

**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 bothÂ sides of AB.
- Let these arcs intersect each other at points M and N.Â 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 7Â·5 cm, the difference of the other two sides is 2Â·5 cm, andÂ one base angle of 45Â°.

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

__Required:__To construct the DABC.Â

**Case (i)**AB â€“ AC = 2Â·5 cm.Â

__Steps of Construction:__Â

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

**Case (ii)**AC â€“ AB = 2Â·5 cm.Â

__Steps of Construction:__

- Draw BC = 7Â·5 cm.Â
- At B, construct âˆ CBX = 45Â° and produce XB toÂ form a line XBXâ€™.
- From BXâ€™, cut off BDâ€™ = 2Â·5 cm.
- Join CDâ€™.Â
- Draw the perpendicular bisector RS of CDâ€™Â 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 perimeterÂ 11Â·8 cm and base angles 60Â° and 45Â°.

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

__Construction:__To construct the DABC.

__Steps of Construction:__Â

- Draw DE = 11Â·8 cm.Â
- At D, construct âˆ EDP = 1/2 of 60Â° = 30Â° and at E, construct âˆ DEQ = 1/2 of 45Â° = 22Â½Â°.Â
- Let DP and EQ meet at A.Â 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.Â