NCERT Solutions for Class 8 Maths Chapter 10 Exercise 10.3 in Hindi and English Medium updated for CBSE session 2022-2023.
Class 8 Maths Chapter 10 Exercise 10.3 Solution
Class VIII Maths Ex. 10.3 Visualising Solid Shapes in PDF file format free to use online as well as offline for CBSE session 2022-23. If you are getting some difficulty to understand particular question, use videos solutions to clear your doubts. Class 8 NCERT mathematics exercise 10.3 explanation and solution videos are given here, explaining all the questions properly.
|Chapter: 10||Exercise: 10.3|
|Topic Name:||Visualising Solid Shapes|
|Content:||NCERT Solution and Explanation|
|Medium:||Hindi and English Medium|
Euler’s Relation for 3-Dimensional Figures
In a 3-dimnsional figure, let the number of faces be F; the number of edges be E and the number or vertices be V.
Then, the Euler’s relation is given by
F – E + V = 2.
Verification of Euler’s Relation for Various Figures:
A Cube or A Cuboid
Number of faces = 6
Number of edges = 12
Number of vertices = 8
F = 5, E = 9 and V= 8
So, (F – E + V) = (6 – 12 + 8) =2
Number of faces = 2, triangular + 3 rectangular = 5
Number of edges = 9
Number of vertices = 6
So, F = 5, E = 9 and V= 6 (F – E + V) = (5 – 9 + 6) = 2
Number of faces = 2 squares + 4 rectangles = 6
Number of edges = 12.
Number of vertices = 8.
So, F = 6, E = 12 and V= 8 (F – E + V) = (6 – 12 + 8) = 2
Number of faces = 2 pentagons + 5 rectangles = 7
Number of edges = 15
Number of vertices = 10.
So, F = 7, E = 15 and V= 10 (F – E + V) = (7- 15 + 10) = 2
A right prism having each of the bases a polygon of n-sides
Number of faces = 2 polygons of n sides + n rectangles = (n + 2)
Number of edges = 3n
Number of vertices = 2n
So, F = (n + 2), E = 3 and V= 2n (F – E + V) = (n + 2 – 3n + 2n) = 2
Triangular pyramid (Tetrahedron)
Number of faces = 4
Number of edges = 6
Number of vertices = 4
So, F = 4, E = 6 and V= 4 (F – E + V) = (4 -6 + 4) =2
Number of faces = 1 squares + 4 rectangles = 5
Number of edges = 8
Number of vertices = 5
So, F = 5, E = 8 and V= 5 (F – E + V) = (5 – 8 + 5) = 2
Number of faces = 1 pentagons + 5 rectangles = 6
Number of edges = 10
Number of vertices = 6
So, F = 6, E = 10 and V= 6 (F – E + V) = (6 – 10 + 6) = 2
A Pyramid whose base is a polygon of n-sides
Number of faces =1 polygons of n sides + n rectangles = (n +1).
Number of edges = 2n
Number of vertices = (n + 1)
F = (n + 1), E =2n and V= (n + 1) (F – E + V) = (n + 1) – (2n) + (n + 1)] = 2
How do exercise 10.3 of class 8 math help you to recognise rectangular prism and cuboid?
A cuboid has a square cross-sectional area and a length, that is possibly different from the side of the cross-section. It has 8 vertices, 12 sides, 6 faces. A rectangular prism has a rectangular cross-section. It may not stand vertical, if you make it stand on the cross sectional base.
How do class 8 Maths exercise 10.3 explore the Euler’s formula?
Euler’s formula, Either of two important mathematical theorems of Leonhard Euler. It is written F + V = E + 2, where F is the number of faces, V the number of vertices, and E the number of edges. A cube, for example, has 6 faces, 8 vertices, and 12 edges, and satisfies this formula.
Is Euler’s rule useful for exercise 10.3 of 8th Maths?
Euler’s formula relates the complex exponential to the cosine and sine functions. This formula is the most important tool in AC analysis. It is why electrical engineers need to understand complex numbers.
The theorem states a relation of the number of faces, vertices, and edges of any polyhedron. The Euler’s formula can be written as F + V = E + 2, where F is the equal to the number of faces, V is equal to the number of vertices, and E is equal to the number of edges.