NCERT Solutions Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

NCERT Solutions Class 7 Mathematics 
Chapter – 10 (Practical Geometry)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 10 Practical Geometry  Exercise 10.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 10 Practical Geometry

• NCERT Solution Class 7 Maths Exercise – 10.1
• NCERT Solution Class 7 Maths Exercise – 10.2
• NCERT Solution Class 7 Maths Exercise – 10.3
• NCERT Solution Class 7 Maths Exercise – 10.5

Exercise – 10.4 

1. Construct ΔABC, given m ∠A =60^o, m ∠B = 30^o and AB = 5.8 cm.

Solution –

Steps of construction:
1. Draw a line segment AB = 5.8 cm.
2. Draw ∠A = 60° and ∠B = 30° to meet each other at C.
Then, ΔABC is the required triangle.

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

Solution –

We know that the sum of the angles of a triangle is 180^o.
∴ ∠PQR + ∠QRP + ∠RPQ = 180^o
⇒ 40^o+ 105^o+ ∠RPQ = 180^o
⇒ 145^o + ∠RPQ = 180^o
⇒ ∠RPQ = 180^o– 145⁰
⇒ ∠RPQ = 35^o
Hence, the measures of ∠RPQ is 35^o.

Steps of construction:
1. Draw a line segment PQ = 5 cm.
2. Draw ∠QPR = 35° and ∠PQR = 40° to meet each other at R.
Then, ΔPQR is the required triangle.

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

Solution – 

To construct ∆DEF with the given measurement, is not possible.
∵ m∠E + m∠F = 110° + 80°
= 190° > 180° [Sum of the angles of a triangle = 180°]
∴ ΔDEF is not possible to construct.