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

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

NCERT Solutions for Class 7 Maths Exercise 10.3 Chapter 10 Practical Geometry in simple PDF are given here. This exercise of NCERT Solutions for Maths 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 subject experts have prepared these NCERT Solutions for Class 7 Maths, Chapter 10 Practical Geometry to help students to clear their doubts.

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

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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 10.4 Solutions

Exercise 10.5 Solutions

Access Answers to NCERT Class 7 Maths 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 make 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 are 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 make 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 make an angle of 60o, i.e., ∠XCB = 60o.

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

4. Join AB.

Then, ΔABC is the required triangle.

Also, explore – 

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 

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