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

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4, are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4.

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

Question 1.

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

Solution:

Steps of Construction

1. Draw AB of length 5.8 cm.
   
   
2. At A, draw a ray AP making an angle of 60° with AB.
3. At B, draw a ray BQ making an angle of 30° with BA.
4. 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

1. Draw PQ of length 5 cm.
2. At Q, draw a ray QX making an angle of 105° with QP.
   
   
3. At P draw a ray PY making an angle of 35° with PQ.
4. 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.

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