In this chapter, we provide RD Sharma Class 6 ex 18.1 Solutions Chapter 18 Basic Geometrical Tools for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 18.1 Solutions Chapter 18 Basic Geometrical Tools Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 6 |

Subject | Maths |

Chapter | Chapter 18 |

Chapter Name | Basic Geometrical Tools |

Exercise | Ex 18.1 |

**RD Sharma Solutions for Class 6 Chapter 18 Basic Geometrical Tools** **Ex 18.1 Download PDF**

**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°

**All Chapter RD Sharma Solutions For Class 6 Maths**

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