# Ncert Solutions For Class 8 Maths Ex 10.3

## Ncert Solutions For Class 8 Maths Chapter 10 Ex 10.3

Question 1:

How many in number should a polygon have got for its face?

(i) Threetriangles

(ii) four triangles

(iii) A square & 4 triangles

(i)Not possible, a polyhedron do not have 3 triangles for its having no faces.

(ii)Yes, a polyhedron can also have 4 triangles which is also known as pyramid which is on a triangular base.

(iii)Yes, a polyhedron has the faces with a square and 4 triangles which makes a pyramid on its square base.

Question 2:

Can any number of faces be added to a polyhedron (Hint: Think of a pyramid)

Yes, if only the number of faces is equal or more than 4.

Question 3:

Which of the following is prism?

Figure (2) unsharpened pencil and (3) a box are the prisms.

Question 4:

Give reasons:

(i)Cylinders and prisms are same and alike.

(ii)Cones and pyramid are same and alike.

(i)If the number of the sides in the prisms become much larger than the prisms tend to become cylinders.

(ii)When the number of the sides of the base increases then the pyramid tend to become a cone.

Question 5:

Is square prism is equal to cube? Give reasons.

No, it is not possible; it can also be a shaped cuboid.

Question6:

Give an adequate response of Euler’s formula for the given solids:

(i)Here, figure (i) Itcontains seven faces, ten vertices and fifteen edges.

Using Euler’s formula,

We tend to seeA + B

C = 2

Putting A = 7, B = 10 and C = 15,

A + B–C = 2

7 + 10–5 = 2

17 –15 = 2

2 = 2

L.H.S. = R.H.S.

(ii)Here, figure (ii) contains 9 faces, 9 Vertices and 16 edges.

Using Euler’s formula, we see

A + B–C = 2

A + B –C = 2

9 + 9 –16 = 2

18 –16 = 2

2 = 2

L.H.S. = R.H.S.

Question7:

Using Euler’s formula, find the unknown:

 Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?

In first column,

A= ?, B = 6 and C = 12

Using Euler’s formula, we see A + B –C = 2

A + B –C = 2

A + 6 –12 = 2

A –6 = 2

A = 2 + 6 = 8

Hence there are 8 faces.

In second column,

A = 5, B= ? and C = 9

Using Euler’s formula, we see A + B –C = 2

A + B –C = 2

5 + B –9 = 2

B –4 = 2

B = 2 + 4 = 6

Hence there are 6 vertices.

In third column,

A = 20, B = 12 and C= ?

Using Euler’s formula, we see A + B–C= 2

A + B –C = 2

20 + 12 –C = 2

32 –C = 2

C = 32 –2 = 30

Hence there are 30 edges.

Question 8:

Can a polyhedron have 10 faces, 20 edges and 15 vertices?

If A = 10, B= 1

5 and C = 20.

Then, we know Using Euler’s formula, A+ B –C = 2

L.H.S.

= A + B –C = 10 + 15–20 = 25 –20 = 5

R.H.S. = 2

L.H.S. $\neq$ R.H.S.

Therefore, it does not follow Euler’s formula