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

*According to the latest CBSE Syllabus 2023-24, this chapter has been removed.

NCERT Solutions for Class 7 Maths Exercise 10.4 Chapter 10 Practical Geometry in simple PDF are given here. Constructing a triangle when the measures of two of its angles and the length of the side inclined between them is given. (ASA Criterion) is the only topic covered in this exercise of NCERT Maths Solutions for Class 7 Chapter 10. Students gain more knowledge by referring to NCERT Solutions for Class 7 Maths, Chapter 10 Practical Geometry.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry – Exercise 10.4

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Access Other Exercises of NCERT Solutions For Class 7 Maths Chapter 10 – Practical Geometry

Exercise 10.1 Solutions

Exercise 10.2 Solutions

Exercise 10.3 Solutions

Exercise 10.5 Solutions

Access Answers to NCERT Class 7 Maths Chapter 10 – Practical Geometry 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. At point A, draw a ray P to making an angle of 60^o i.e. ∠PAB = 60^o.

3. At point B, draw a ray Q to making an angle of 30^o i.e. ∠QBA = 30^o.

4. Now the two rays AP and BQ intersect at the point 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

= 105^o+ 40^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. At point P, draw a ray L to making an angle of 105^o i.e. ∠LPQ = 35^o.

3. At point Q, draw a ray M to making an angle of 40^o i.e. ∠MQP = 105^o.

4. Now the two rays PL and QM intersect at the point 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:-

From the question it is given that,

EF = 7.2 cm

∠E = 110^o

∠F = 80^o

Now we have to check whether it is possible to construct ΔDEF from the given values.

We know that the sum of the angles of a triangle is 180^o.

Then,

∠D + ∠E + ∠F = 180^o

∠D + 110^o+ 80^o= 180^o

∠D + 190^o = 180^o

∠D = 180^o– 190⁰

∠D = -10^o

We may observe that the sum of two angles is 190^o is greater than 180^o. So, it is not possible to construct a triangle.

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NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 

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