NCERT Solutions for Class 7 Maths Exercise 10.3 Chapter 10 Practical Geometry

NCERT Solutions For Class 7 Maths Ex 10.3 PDF Free Download

NCERT Solutions for Class 7 Maths Exercise 10.3 Chapter 10 Practical Geometry in simple PDF are given here. This exercise of NCERT Solutions for Class 7 Chapter 10 contains topics related to constructing a triangle when the lengths of two sides and the measure of the angle between them are known (SAS Criterion). Our expert personnel have prepared these NCERT Solutions for Class 7 Maths, Chapter 10 Practical Geometry to help students to clear their doubts.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 10 Practical Geometry – Exercise 10.3

 

ncert sol class 7 math ch 10 ex 3
ncert sol class 7 math ch 10 ex 3

 

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

Exercise 10.1 Solutions

Exercise 10.2 Solutions

Exercise 10.4 Solutions

Exercise 10.5 Solutions

Access answers to Maths NCERT Solutions for Class 7 Chapter 10 – Practical Geometry Exercise 10.3

1. Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90o.

Solution:-

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Image 8

Steps of construction:

1. Draw a line segment DF = 3 cm.

2. At point D, draw a ray DX to making an angle of 90o i.e. ∠XDF = 90o.

3. Along DX, set off DE = 5cm.

4. Join EF.

Then, ΔEDF is the required right angled triangle.

2. Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110o.

Solution:-

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Image 9

Steps of construction:

1. Draw a line segment AB = 6.5 cm.

2. At point A, draw a ray AX to making an angle of 110o i.e. ∠XAB = 110o.

3. Along AX, set off AC = 6.5cm.

4. Join CB.

Then, ΔABC is the required isosceles triangle.

3. Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

Solution:-

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

Steps of construction:

1. Draw a line segment BC = 7.5 cm.

2. At point C, draw a ray CX to making an angle of 60o i.e. ∠XCB = 60o.

3. Along CX, set off AC = 5cm.

4. Join AB.

Then, ΔABC is the required triangle.


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