**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) 90 ^{o} (b) 45^{o}**

**Solution:**

**(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 – 90^{o}

**(b)** 45^{o} angle using compass

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

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

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

∠AOB = 45^{o}

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

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

**Solution:**

(a) 30^{o} 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 60^{o} to AB

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

**(e)** 105^{o} using compass

Step 1: – Draw a 90^{o }angle

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

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

∠CAB = 105^{o}

**(f)** 135^{o} using compass

Step 1: – Draw 120^{o}

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

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

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

∠CAB = 135^{o}

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

**Solution:**

Equilateral triangle has equal angles and equal side

∴ Each angle = 60^{o} [∵3 x 60 = 180^{o}]

Each side = 4.5 cm [given]

(Picture)

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 60^{o}. the intersection is the 3^{rd} 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]**

**Solution:**

Draw two 45^{o} angle on each end of line segment AB of any length and joint the two lines at 45^{o} 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 = 45^{o}

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