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Class 9 Mathematics Revision Notes for Constructions of Chapter 11
Chapter 11 of the NCERT book for Class 9 Mathematics is titled ‘Constructions.’ Students will learn about the approach for precisely building angles and various 2D shapes of the relevant dimensions in this chapter. Students would learn how to use a compass, a protractor, dividers, and set squares as part of the procedure.
Generally, the chapter is viewed as comfortable by students. However, a student must understand the techniques and logic involved in construction to score better in the examination. Proper comprehension and practice of all NCERT questions in a stepbystep and thorough way are required.
Class 9 Mathematics Revision Notes for Constructions of Chapter 11
Access Class 9 Mathematics Chapter 11 – Constructions Notes
Introduction:
In geometrical constructions, drawing is done solely with a nongraduated ruler known as a straightedge and a compass. A graded scale and a protractor may also be used when measurements are necessary.
Basic constructions:
To construct the bisector of a given angle, start with an ∠ABC and bisect it. The steps that are to be taken are as follows:
 Draw an arc of any radius with B as its centre, intersecting the rays BA and BC at E and D, respectively.
 Draw arcs with E and D as centres and a radius greater than 12DE so that they intersect at F.
 Draw a ray connecting B and F and extend it. As a result, BF is the required angle bisector.
To draw the perpendicular bisector of a given line segment – Assume a line segment AB as a perpendicular bisector. The actions to take are as follows:
 Using A and B as the centres, draw arcs with a radius greater than 12AB on both sides of the line segment. The drawn arcs should intersect each other.
 Assume that the arcs intersect at P and Q, respectively. P and Q are to be joined.
 Assume that the line PQ crosses segment AB at point M. The required perpendicular bisector is then the line PMQ.
To create an angle of 60о at the beginning of a given ray, start with a ray AB with an initial point A from which we must draw another ray AC so that ∠BAC=60о. The actions to take are as follows:
 Using A as the centre, draw an arc that cuts the ray AB at point D. Now, with A and D as the centres and AD as the radius, draw two arcs that intersect at point E.
 Join AE and extend it all the way to point C on the ray AC. As a result, ∠CAB=60о.
Some Construction of Triangles:
For the construction of a triangle, measurements of at least three triangle elements are necessary. However, all threeelement combinations are insufficient for the goal. For example, it isn’t possible to construct a completely unique triangle while the measurements of aspects and a perspective that isn’t protected in among the given aspects are offered. A triangle can be constructed when, (i) the base, one base attitude, and the total of the opposite aspects are specified; (ii) the base, a base attitude, and the distinction between opposed aspects are provided; (iii) The perimeter and base angles are specified.
To construct a triangle, given its base, a base angle, and the sum of its other two sides – say we have to draw ΔABC, whose base BC and ∠B are given. In addition, the sum AB+AC is given. The required triangle will be built in the following steps:
 Draw the base BC first, and then make ∠CBX equal to the angle provided at point B.
 Make a mark D on the ray BX so that BD=AB+AC.
 Then join DC. Construct ∠DCY in such a way that it equals ∠BDC.
 Assume that ray CY intersects BD at point A. The required triangle is then ΔABC.
To construct a triangle given its base, a base angle, and the difference of the other two sides – say, ΔABC, of which the base BC, one base angle ∠B, and the difference of the other two sides ABAC or ACAB is given. There are two possibilities here:
Case 1: When AB>AC
 Draw the base BC first, and then make ∠CBX equal to the angle given to us at point B.
 Make a mark D on the ray BX so that BD=ABAC.
 The ray BX is cut at A by the perpendicular bisector of line CD, denoted by PQ when extended. As a result, ΔABC has been constructed.
Case (ii): When AB<AC
 Draw the base BC first, and then make ∠CBX equal to the angle given to us at point B.
 Mark a point D on the ray BX that extends on the other side of BX so that BD=ACAB.
 The ray BX is cut at A by the perpendicular bisector of line CD, denoted by PQ when extended. As a result, ΔABC has been built.
To construct a triangle, given its perimeter and its two base angles – Construct ΔABC where the base angles ∠B and ∠C and the perimeter AB+BC+CA are given. The steps that must be followed are:
 Draw a line XY that equals AB+BC+CA.
 Then make ∠LXY=∠B and ∠MYX=∠C.
 Divide the angles ∠LXY and ∠MYX in half so that the bisectors meet at point A.
 Construct perpendicular bisectors PQ and RS from AX and AY, respectively.
 Assume PQ intersects XY at B and RS intersects XY at C. After that, join AB and AC. As a result, we get the needed ΔABC.
FAQs (Frequently Asked Questions)
1. What do we learn in the chapter Constructions?
This chapter on Constructions is designed to familiarise pupils with compasses and rulers. Students can learn how to divide an angle in half, to create a perpendicular bisector for the purpose of bisecting a line segment, to create angles with specific degrees such as 30°, 60°, 90°, and so on, to create a triangle, provide its perimeter and two base angles, and to build a triangle with a specified base angle and calculate the remaining two angles.
2. How to construct an equilateral triangle with a given side.
It is known that an equilateral triangle has equal sides as well as equal angles of 60o each. As a result, all sides of the equilateral triangle will be 5cm long. To draw an equilateral triangle of 5cm side, the instructions below will be used.
 Step I: Draw a 5cmlong line segment AB. Draw an arc of some radius with A as its centre. Allow it to intersect AB at P.
 Step II: With P as the centre, draw an arc that intersects the preceding arc at E. Join the AE team.
 Step III: With A as the centre, draw a 5cm radius arc that crosses the extended line segment AE at C. Accompany AC and BC. ΔABC is the required equilateral triangle of side 5cm.