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
- 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.
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