Chapter 18: Basic Geometrical Tools Exercise 18.1

Question: 1

Construct the following angles using set- squares:

(i) 45^°

(ii) 90^°

(iii) 60^°

(iv) 105^°

(v) 75^°

(vi) 150^°

Solution:

(i) 45^°

Place 45^° set- square.

Draw two rays AB and AC along the edges from the vertex from the vertex of 45^o angle of the set- square.

The angle so formed is a 45^° angle.

∠BAC = 45°

(ii) 90°

Place = 90^° set –square as shown in the figure.

Draw two rays BC and BA along the edges from the vertex of 90^° angle.

The angle so formed is 90^° angle.

∠ABC = 90°

(iii) 60^°

Place 30^° set –square as shown in the figure.

Draw the rays BA and BC along the edges from the vertex of 60^°

The angle so formed is 60^°

∠ABC = 60°

(iv) 105^°

Place 30^° set –square and make an angle 60^° by drawing the rays BA and BC as shown in figure.

Now place the vertex of 45^°of the set –square on the ray BA as shown in figure and draw the ray BD.

The angle so formed is 105^°

Therefore, ∠DBC = 105°

(v) 75^o

Place 45^° set –square and make an angle of 45^° by drawing the rays BD and BC as shown in the figure.

Now place the vertex of 30^° of the set- square on the ray BD as shown in the figure and draw the ray BA.

The angle so formed is 75^°.

Therefore, ∠ABC = 75°

(Line BD is hidden)

(vi) 150^°

Place the vertex of 45^° of the set – square and make angle of 90^o by drawing the rays BD and BC as shown in the figure

Now, place the vertex of 30^°of the set –square on the ray BS as shown in the figure and draw the ray BA

The angle so formed is 150^°.

Therefore, ∠ABC = 150°

Question: 2

Given a line BC and a point A on it, construct a ray AD using set – squares so that ∠DAC is

(i) 30^°

(ii) 150^°

Solution:

(i) Draw a line BC and take a point A on it. Place 30^° set –square on the line BC such that its vertex of 30^° angle lies on point A and one edge coincides with the ray AB as shown in figure

Draw the ray AD.

Thus ∠DAC is the required angle of 30^°

(ii) Draw a line BC and take a point A on it. Place 30^° set –square on the line BC such that its vertex of 30^° angle lies on point A and one edge coincides with the ray AB as shown in the figure.

Draw the ray AD.

Therefore, ∠DAB = 30°

We know that angle on one side of the straight line will always add to 180^o

Therefore, ∠DAB + ∠DAC = 180°

Therefore, ∠DAC = 150°