# RD Sharma Solutions Class 6 Basic Geometrical Tools Exercise 18.1

## RD Sharma Solutions Class 6 Chapter 18 Exercise 18.1

Exercise 18.1

1.) Construct the following angles using set- squares:

(i) 45o

Place 45o set- square.

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

The angle so formed is a 45 o angle.

$\angle BAC = 45^{o}$

(ii) 90o

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

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

The angle so formed is 90o angle.

$\angle ABC = 90^{o}$

(iii) 60o

Place 30o set –square as shown in the figure.

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

The angle so formed is 60o

$\angle ABC = 60^{o}$

(iv) 105o

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

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

The angle so formed is 105o

Therefore, $\angle DBC = 105^{o}$

(v ) 75o

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

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

The angle so formed is 75o.

Therefore, $\angle ABC = 75^{o}$

(Line BD is hidden)

(vi) 150o

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

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

The angle so formed is 150o.

Therefore, $\angle ABC = 150^{o}$

2.) Given a line BC and a point A on it, construct a ray AD using set – squares so that $\angle DAC$ is

(i) 30o

(ii) 150o

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

Thus $\angle DAC$ is the required angle of 30o

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

Therefore, $\angle DAB = 30^{o}$
Therefore, $\angle DAB + \angle DAC = 180 ^{o}$
Therefore, $\angle DAC = 150 ^{o}$