NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3.

 Board CBSE Textbook NCERT Class Class 8 Subject Maths Chapter Chapter 10 Chapter Name Visualising Solid Shapes Exercise Ex 10.3 Number of Questions Solved 8 Category NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3

Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution.
(i) No
(ii) Yes
(iii) Yes

Question 2.
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Solution.
Possible, only if the number of faces is greater than or equal to 4.

Question 3.
Which are prisms among the following?

Solution.
We know that a prism is a polyhedron whose base and top faces are congruent and parallel and other (lateral) faces are parallelograms in shape. So, only (ii) and (iv) are prisms.

Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution.
(i) The prisms and cylinder, both, have their base and top faces as congruent and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.

(ii) The pyramids and cones are alike in the sense that their lateral faces meet at a point (called vertex). Also, a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Question 5.
Is a square prism same as a cube ? Explain.
Solution.
No; not always as it can be a cuboid also.

Question 6.
Verify Euler’s formula for these solids

Solution.
(i)
F = 7
V= 10
E = 15
F + V = 7 + 10 = 17
E + 2 = 15 + 2 = 17
So, F + V = E + 2
Hence, Euler’s Formula is verified,

(ii)
F = 9
V = 9
E = 16
F + V = 9 + 9 = 18
E + 2 = 16 + 2 = 18
So, F + V = E + 2
Hence, Euler’s Formula is verified

Question 7.
Using Euler’s formula find the unknown.

Solution.
(i)
F + V = E + 2
⇒ F + 6 = 12 + 2
⇒ F + 6 = 14
⇒ F = 14 – 6 = 8

(ii)
F + V = E + 2
⇒ 5 + V = 9 + 2
⇒ 5 + V = 11
⇒ V= 11-5 = 6

(iii)
F + V = E + 2
⇒ 20 + 12 = E + 2
⇒ 32 = E + 2
⇒ E = 32 – 2
⇒ E = 30

Question 8.
Can a polyhedron have 10 faces, 20 edges, and 15 vertices?
Solution.
Here F = 10
E = 20
V= 15
So, F + V = 10 + 15 = 25
E + 2 = 20 + 2 = 22
∵ F + V ≠ E + 2
∴ Such a polyhedron is not possible.

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