Telangana SCERT Class 9 Math Solution Chapter 13 Geometrical Constructions Exercise 13.1

Telangana SCERT Solution Class IX (9) Math Chapter 13 Geometrical Constructions Exercise 13.1

(1.) Construct the following angles at the initial point of a given ray and justify the construction.

(a) 90o (b) 45o


(a) 90 o angle using compass

Step 1: – Draw a line segment AB

Step 2: – From point A as centre draw an arc that cut AB at a point

Step 3: – With same radius cut the arc taking centre as the previous point where the arc cut AB

Step 4: – From this new point take centre with same radius and cut the arc once again.

Step 5: – From this two point on the arc take two arc that cuts each other.

Step 6: – Join the point with A using a ruler ∠CAB – 90o

(b) 45o angle using compass

Step 1: – Draw a 90o angle using the process shown in the previous sum.

Step 2: – Bisect  the 90o angle by taking any radius and drawing two arcs on the 90o plane from 1st point if the base whose the 1st arc cuts and the point where the perpendicular cuts the 1st one

Step 3: – joint the point O and the instruction point of the two arcs.

∠AOB = 45o

(2.) Construct the following angles using ruler and compass and verify by measuring them by a protractor.

(a) 30o (b) 22 ½ o (c) 15o (d) 75o (e) 105o f) 135o


(a) 30o angles using compass

Step 1: – Draw AB line segment.

Step 2: – From A as centre draw an arc which cuts base AB.

Step 3; – From this point in base AB taking same radii cut the same arc. The line joining A & this point is 60o to AB

Step 4: – Bisect the 60o angle as show in sum 1. <CAB = 30o

(e) 105o using compass

Step 1: – Draw a 90o angle

Step 2: – Line passing though point O and B is 120o

Step 3: – Bisect the angle between 90o & 120o

∠CAB = 105o

(f) 135o using compass

Step 1: – Draw 120o

Step 2: – From 120 arc as centre with same radii out the 1st arc which is 180o

Step 3: – Bisect angle between 120o & 180o you get 150o

Step 4: – Bisect the angle between 150o & 120o you get 135o

∠CAB = 135o

(3.) Construct an equilateral triangle, given its side of length of 4.5 cm and justify the construction.


Equilateral triangle has equal angles and equal side

∴ Each angle = 60o [∵3 x 60 = 180o]

Each side = 4.5 cm [given]


Step 1: – Draw a line segment AB = 4.5cm

Step 2: – Draw two 60 angle at point A & B

Step 3: – Join those two lines from A and B at 60o. the intersection is the 3rd vertex.

Hence, △ABC is an Equilateral triangle.

(4.) Construct an isosceles triangle, given its base and base angle and justify the construction. [Hint : You can take any measure of side and angle]


Draw two 45o angle on each end of line segment AB of any length and joint the two lines at 45o angle we get △ACB which is an isosceles triangle

Isosceles triangle has two equal angles and two equal sides. Here AB is the base AC and BC are two equal sided they are opposite to equal angles <CAB = <CBA = 45o

You can take any acute angle except 60o angle since two 60o angle will make equilateral angle.

Updated: October 5, 2021 — 4:25 pm

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