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Basic Geometrical Tools (Exercise 18.1) RD Sharma Solutions | Mathematics (Maths) Class 6 PDF Download

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 Page 1


 
 
 
 
 
 
                                                                              
1. Construct the following angles using set-squares: 
(i) 45
0
 
(ii) 90
0
 
(iii) 60
0
 
(iv) 105
0
 
(v) 75
0
 
(vi) 150
0
 
Solution: 
 
(i) Now place 45
0
 using set square. Construct two rays AB and AC along the edges from the vertex A of 45
0
 angle 
of the set square. So the angle formed is 45
0
 
Hence, ?BAC = 45
0
 
 
(ii) Now place 90
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 90
0
 
angle of the set square. So the angle formed is 90
0
 
Hence, ?ABC = 90
0
 
g 
(iii) Now place 60
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 60
0
 
angle of the set square. So the angle formed is 60
0
 
Hence, ?ABC = 60
0
 
Page 2


 
 
 
 
 
 
                                                                              
1. Construct the following angles using set-squares: 
(i) 45
0
 
(ii) 90
0
 
(iii) 60
0
 
(iv) 105
0
 
(v) 75
0
 
(vi) 150
0
 
Solution: 
 
(i) Now place 45
0
 using set square. Construct two rays AB and AC along the edges from the vertex A of 45
0
 angle 
of the set square. So the angle formed is 45
0
 
Hence, ?BAC = 45
0
 
 
(ii) Now place 90
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 90
0
 
angle of the set square. So the angle formed is 90
0
 
Hence, ?ABC = 90
0
 
g 
(iii) Now place 60
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 60
0
 
angle of the set square. So the angle formed is 60
0
 
Hence, ?ABC = 60
0
 
 
 
 
 
 
 
 
(iv) Now place 30
0
 using set square and make an angle 60
0
. Construct two rays BA and BC as given in figure. 
 
Place the set square of angle 45
0
 of the set-square on the ray BA and draw the ray BD. So the angle formed is 
105
0
.  
Hence, ?DBC = 105
0
 
 
(v) Now place 45
0
 using set square and make an angle 45
0
. Construct two rays BC and BD as shown in figure. 
Page 3


 
 
 
 
 
 
                                                                              
1. Construct the following angles using set-squares: 
(i) 45
0
 
(ii) 90
0
 
(iii) 60
0
 
(iv) 105
0
 
(v) 75
0
 
(vi) 150
0
 
Solution: 
 
(i) Now place 45
0
 using set square. Construct two rays AB and AC along the edges from the vertex A of 45
0
 angle 
of the set square. So the angle formed is 45
0
 
Hence, ?BAC = 45
0
 
 
(ii) Now place 90
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 90
0
 
angle of the set square. So the angle formed is 90
0
 
Hence, ?ABC = 90
0
 
g 
(iii) Now place 60
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 60
0
 
angle of the set square. So the angle formed is 60
0
 
Hence, ?ABC = 60
0
 
 
 
 
 
 
 
 
(iv) Now place 30
0
 using set square and make an angle 60
0
. Construct two rays BA and BC as given in figure. 
 
Place the set square of angle 45
0
 of the set-square on the ray BA and draw the ray BD. So the angle formed is 
105
0
.  
Hence, ?DBC = 105
0
 
 
(v) Now place 45
0
 using set square and make an angle 45
0
. Construct two rays BC and BD as shown in figure. 
 
 
 
 
 
 
 
Place the vertex of 30
0
 of the set square on the ray BD and draw the ray BA. So the angle formed is 75
0
 
Hence, ?ABC = 75
0
.  
 
(vi) Now place the vertex of 45
0
 of the set square and make an angle 90
0
 and construct two rays BD and BC as 
shown in figure.  
 
Place the vertex of 30
0
 of the set square on the ray BC and draw the ray BA. So the angle formed is 150
0
 
Page 4


 
 
 
 
 
 
                                                                              
1. Construct the following angles using set-squares: 
(i) 45
0
 
(ii) 90
0
 
(iii) 60
0
 
(iv) 105
0
 
(v) 75
0
 
(vi) 150
0
 
Solution: 
 
(i) Now place 45
0
 using set square. Construct two rays AB and AC along the edges from the vertex A of 45
0
 angle 
of the set square. So the angle formed is 45
0
 
Hence, ?BAC = 45
0
 
 
(ii) Now place 90
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 90
0
 
angle of the set square. So the angle formed is 90
0
 
Hence, ?ABC = 90
0
 
g 
(iii) Now place 60
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 60
0
 
angle of the set square. So the angle formed is 60
0
 
Hence, ?ABC = 60
0
 
 
 
 
 
 
 
 
(iv) Now place 30
0
 using set square and make an angle 60
0
. Construct two rays BA and BC as given in figure. 
 
Place the set square of angle 45
0
 of the set-square on the ray BA and draw the ray BD. So the angle formed is 
105
0
.  
Hence, ?DBC = 105
0
 
 
(v) Now place 45
0
 using set square and make an angle 45
0
. Construct two rays BC and BD as shown in figure. 
 
 
 
 
 
 
 
Place the vertex of 30
0
 of the set square on the ray BD and draw the ray BA. So the angle formed is 75
0
 
Hence, ?ABC = 75
0
.  
 
(vi) Now place the vertex of 45
0
 of the set square and make an angle 90
0
 and construct two rays BD and BC as 
shown in figure.  
 
Place the vertex of 30
0
 of the set square on the ray BC and draw the ray BA. So the angle formed is 150
0
 
 
 
 
 
 
 
 
Hence, ?ABC = 150
0
 
2. Given a line BC and a point A on it, constructed a ray AD using set squares so that ?DAC is  
(i) 30
0
 
(ii) 150
0
 
Solution: 
 
(i) Construct a line BC and mark a point A on it. Now place a 30
0
 set square on the line BC where its vertex of 30
0
 
angle lies on point A and one edge coincides with the ray AB as shown in figure. Construct the ray AD. 
 
Hence, the required ?DAC = 30
0
 
 
(ii) Construct a line BC and mark a point A on it. Now place 30
0
 set square on the line BC where its vertex of 30
0
 
lies on point A and one edge coincides with the ray AB as shown in figure. Construct the ray AD 
Hence, ?DAB = 30
0
 
 
Page 5


 
 
 
 
 
 
                                                                              
1. Construct the following angles using set-squares: 
(i) 45
0
 
(ii) 90
0
 
(iii) 60
0
 
(iv) 105
0
 
(v) 75
0
 
(vi) 150
0
 
Solution: 
 
(i) Now place 45
0
 using set square. Construct two rays AB and AC along the edges from the vertex A of 45
0
 angle 
of the set square. So the angle formed is 45
0
 
Hence, ?BAC = 45
0
 
 
(ii) Now place 90
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 90
0
 
angle of the set square. So the angle formed is 90
0
 
Hence, ?ABC = 90
0
 
g 
(iii) Now place 60
0
 using set square. Construct two rays BC and BA along the edges from the vertex B of 60
0
 
angle of the set square. So the angle formed is 60
0
 
Hence, ?ABC = 60
0
 
 
 
 
 
 
 
 
(iv) Now place 30
0
 using set square and make an angle 60
0
. Construct two rays BA and BC as given in figure. 
 
Place the set square of angle 45
0
 of the set-square on the ray BA and draw the ray BD. So the angle formed is 
105
0
.  
Hence, ?DBC = 105
0
 
 
(v) Now place 45
0
 using set square and make an angle 45
0
. Construct two rays BC and BD as shown in figure. 
 
 
 
 
 
 
 
Place the vertex of 30
0
 of the set square on the ray BD and draw the ray BA. So the angle formed is 75
0
 
Hence, ?ABC = 75
0
.  
 
(vi) Now place the vertex of 45
0
 of the set square and make an angle 90
0
 and construct two rays BD and BC as 
shown in figure.  
 
Place the vertex of 30
0
 of the set square on the ray BC and draw the ray BA. So the angle formed is 150
0
 
 
 
 
 
 
 
 
Hence, ?ABC = 150
0
 
2. Given a line BC and a point A on it, constructed a ray AD using set squares so that ?DAC is  
(i) 30
0
 
(ii) 150
0
 
Solution: 
 
(i) Construct a line BC and mark a point A on it. Now place a 30
0
 set square on the line BC where its vertex of 30
0
 
angle lies on point A and one edge coincides with the ray AB as shown in figure. Construct the ray AD. 
 
Hence, the required ?DAC = 30
0
 
 
(ii) Construct a line BC and mark a point A on it. Now place 30
0
 set square on the line BC where its vertex of 30
0
 
lies on point A and one edge coincides with the ray AB as shown in figure. Construct the ray AD 
Hence, ?DAB = 30
0
 
 
 
 
 
 
 
 
We know that the angle on one side of the straight line will always add to 180
0
 
So we get ?DAB + ?DAC = 180
0
 
Hence, ?DAC = 150
0
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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