Ncert Solutions For Class 7 Maths Ex 10.4

NCERT Solutions For Class 7 Maths Ex 10.4 PDF Free Download

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 Solutions for Class 7 Chapter 10. Students gain more knowledge by referring to NCERT Solutions for Class 7 Maths, Chapter 10 Practical Geometry.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 10 Practical Geometry – Exercise 10.4

 

ncert sol class 7 math ch 10 ex 4
ncert sol class 7 math ch 10 ex 4
ncert sol class 7 math ch 10 ex 4

 

Access other exercises of NCERT Solutions For Class 7 Chapter 10 – Practical Geometry

Exercise 10.1 Solutions

Exercise 10.2 Solutions

Exercise 10.3 Solutions

Exercise 10.5 Solutions

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

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

Solution:-

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Image 11

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 60o i.e. ∠PAB = 60o.

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

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 = 105o and m∠QRP = 40o.

(Hint: Recall angle-sum property of a triangle).

Solution:-

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Image 12

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

∴ ∠PQR + ∠QRP + ∠RPQ = 180o

= 105o+ 40o+ ∠RPQ = 180o

= 145o + ∠RPQ = 180o

= ∠RPQ = 180o– 1450

= ∠RPQ = 35o

Hence, the measures of ∠RPQ is 35o.

Steps of construction:

1. Draw a line segment PQ = 5 cm.

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

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

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 = 110o

∠F = 80o

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 180o.

Then,

∠D + ∠E + ∠F = 180o

∠D + 110o+ 80o= 180o

∠D + 190o = 180o

∠D = 180o– 1900

∠D = -10o

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


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