# NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Here, below you all know about NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 Question Answer. I know many of you confuse about finding this Chapter 10 Practical Geometry Ex 10.4 Of Class 7 NCERT Solutions. So, Read the full post below and get your solutions.

Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 10 |

Chapter Name | Practical Geometry Ex 10.4 |

Number of Questions Solved | 6 |

## NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

**NCERT TEXTBOOK EXERCISES**

**Question 1. Construct ∆ ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.**

**Solution:**

**Steps of Construction**

- Draw AB of length 5.8 cm.
- At A, draw a ray AP making an angle of 60° with AB.
- At B, draw a ray BQ making an angle of 30° with BA.
- Mark the point of intersection of two rays as C.
- ∆ ABC is now completed.

**Question 2. Construct ∆ PQR if PQ = 5 cm, m ∠PQR = 105° and m ∠QRP = 40°.**(Hint: Recall angle-sum property of a triangle).

**Solution:**

By angle-sum property of a triangle

m ∠RPQ + m ∠PQR + m ∠QRP = 180°

⇒ m ∠RPQ + 105° + 40° = 180°

⇒ m ∠RPQ + 145° = 180°

⇒ m ∠RPQ = 35°**Steps of Construction**

- Draw PQ of length 5 cm.
- At Q, draw a ray QX making an angle of 105° with QP.
- At P draw a ray PY making an angle of 35° with PQ.
- Mark the point of intersection of two rays as R.

∆ PQR is now completed.

**Question 3. Examine whether you can construct ∆DEF such that EF = 7.2 cm, m ∠E = 110° and m ∠F = 80°. Justify your answer.**

**Solution:**

m ∠E + m ∠F = 110° + 80° = 190° > 180°

This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of two angles of a triangle cannot exceed 180°.

Hence, ∆ DEF cannot be constructed.

## NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 PDF

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