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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 45o 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) 75o 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 90o 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 180o Therefore, ∠DAB + ∠DAC = 180° Therefore, ∠DAC = 150°
Construct the following angles using set- squares:
(i) 45°
(ii) 90°
(iii) 60°
(iv) 105°
(v) 75°
(vi) 150°
Place 45° 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° angle.
∠BAC = 45°
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°
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°
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) 75o
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)
Place the vertex of 45° 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 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°
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°
(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.
Therefore, ∠DAB = 30°
We know that angle on one side of the straight line will always add to 180o
Therefore, ∠DAB + ∠DAC = 180°
Therefore, ∠DAC = 150°
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Chapter 18: Basic Geometrical Tools Exercise 18.2...
