**Exercise-10.3**

class 8 maths chapter 10 exercise 10.3 (1)

1. Can a polyhedron have for its faces

(i) 3 triangles? (ii) 4 triangles?

(iii) a square and four triangles?

Solution to class 8 maths chapter 10 exercise 10.3 (1)

(i) No, a polyhedron cannot have 3 triangles for its faces.

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

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

class 8 maths chapter 10 exercise 10.3 (2)

2. Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).

Solution to class 8 maths chapter 10 exercise 10.3 (2)

It is possible, only if the number of faces are greater than or equal to 4.

class 8 maths chapter 10 exercise 10.3 (3)

3. Which are prisms among the following?

Solution to class 8 maths chapter 10 exercise 10.3 (3)

Figure (ii) unsharpened pencil

and figure (iv) a box are prisms.

class 8 maths chapter 10 exercise 10.3 (4)

4. (i) How are prisms and cylinders alike?

(ii) How are pyramids and cones alike?

Solution to class 8 maths chapter 10 exercise 10.3 (4)

(i) A prism becomes a cylinder as the number of sides of its base becomes larger and larger

(ii) A pyramid becomes a cone as the number of sides of its base becomes larger and larger.

class 8 maths chapter 10 exercise 10.3 (5)

5. Is a square prism same as a cube? Explain.

Solution to class 8 maths chapter 10 exercise 10.3 (5)

No, a square prism may be cube or cuboid.

class 8 maths chapter 10 exercise 10.3 (6)

6. Verify Euler’s formula for these solids.

Solution to class 8 maths chapter 10 exercise 10.3 (6)

(i)Here, figure (i) contains 7 faces, 10 vertices and 15 edges. Using Euler’s formula, we see F+ V=E+2 Putting F = 7, V = 10 and E = 15,

F+ V=E+2

⇒7+ 10 = 15 +2

⇒17 =17

L.H.S. = RHS.

(ii)Here, figure (ii) contains 9 faces, 9 vertices and 16 edges. Using Euler’s formula, we see F+ V=E+2 Putting F = 9, V = 9 and E = 16,

F+ V=E+2

⇒9+ 9 = 16 +2

⇒18 =18

L.H.S. = RHS.

class 8 maths chapter 10 exercise 10.3 (7)

7. Using Euler’s formula find the unknown.

Solution to class 8 maths chapter 10 exercise 10.3 (7)

In first column F=? V=6 and E=12

Using Euler’s formula, F+ V=E+2

F+ 6=12+2

⇒F+6=14

⇒F=14 –6=8

Hence there are 8 faces.

In second column F=5 V=? and E=9

Using Euler’s formula, F+ V=E+2

5+ V=9+2

⇒V+5=11

⇒V=11 –5=6

Hence there are 6 vertices.

In third column F=20 V=12 and E=?

Using Euler’s formula, F+ V=E+2

20+ 12=E+2

⇒32=E+2

⇒E=32–2=30

Hence there are 30 edges.

class 8 maths chapter 10 exercise 10.3 (8)

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

If Faces=10, Edges=20 and Vertices=15

Then using Euler’s formula, F+ V=E+2

L.H.S=10+15=25

R.H.S=20+2=22

L.H.S≠R.H.S

a polyhedron can not have 10 faces, 20 edges and 15 vertices.

Since it doesn’t follow Euler’s formula.