**4. ** Solution:

**5.** (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

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

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

**7. **Figure (ii) unsharpened pencil and figure (iv) a box are prisms.

**8. **(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.

**9. **No, it can be a cuboid also.

**10. ** (i) Here, figure (i) contains 7 faces, 10 vertices and 15 edges.

Using Eucler’s formula, we see F + V – E = 2

Putting F = 7, V = 10 and E = 15,

F + V – E = 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 Eucler’s formula, we see F + V – E = 2

F + V – E = 2 ⇒ 9 + 9 – 16 = 2

⇒ 18 – 16 = 2 ⇒ 2 = 2

⇒ L.H.S. = R.H.S.

Hence verified Eucler’s formula.

**11. **In first column, F = ?, V = 6 and E = 12

Using Eucler’s formula, we see F + V – E = 2

F + V – E = 2 ⇒ F + 6 – 12 = 2

⇒ F – 6 = 2 ⇒ F = 2 + 6 = 8

Hence there are 8 faces.

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

Using Eucler’s formula, we see F + V – E = 2

F + V – E = 2 ⇒ 5 + V – 9 = 2

⇒ V – 4 = 2 ⇒ V= 2 + 4 = 6

Hence there are 6 vertices.

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

Using Eucler’s formula, we see F + V – E = 2

F + V – E = 2 ⇒ 20 + 12 – E = 2

⇒ 32 – E = 2 ⇒ E = 32 – 2 = 30

Hence there are 30 edges.

**12. ** If F = 10, V = 15 and E = 20.

Then, we know Using Eucler’s formula, F + V – E = 2

L.H.S. = F + V – E

= 10 + 15 – 20

= 25 – 20

=5

R.H.S. = 2

∵ L.H.S. ≠ R.H.S.

Therefore, it does not follow Eucler’s formula.

So polyhedron cannot have 10 faces, 20 edges and 15 vertices.