# Class 8 NCERT Solutions – Chapter 10 Visualising Solid Shapes – Exercise 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) 3 triangles:No, because polyhedron must have minimum 4 faces i.e all edges should meet at vertices.

(ii) 4 triangles:Yes, as all the edges are meeting at the vertices and has four triangular faces.

(iii) a square and four triangles:Yes, because all the eight edges meet at the vertices having a square face and four triangular faces.

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

**Solution:**

Yes, It is possible to have a polyhedron with any given faces only if the number of faces are greater than or equal to four.

**Question 3: **Which are prisms among the following?

**Solution:**

Prisms among the given images are

(ii) Unsharpened pencil

(iv) A box.

**Question 4: **(i) How are prisms and cylinders alike? (ii) How are pyramids and cones alike?

**Solution:**

(i)If the number of sides in a prism are increased to certain extent, then the prism will take the shape of cylinder i.e. a prism with a circular base.

(ii)If the number of sides of the pyramid is increased to same extent, then the pyramid becomes a cone i.e. a pyramid with a circular base.

**Question 5:** Is a square prism same as a cube? Explain.

**Solution: **

Yes, a square prism can be same as a cube, but if the height of the prism is greater than It may be cuboid.

**Question 6:** Verify Eulerâ€™s formula for these solids.

**(i) **

**(ii) **

**Solution:**

(i)No. of Faces (F) = 7No. of Vertices (V) = 10

No. of Edges (E) = 15

By Using Eulerâ€™s formula: F + V â€“ E = 2 and Substituting the values, we get

â‡’ 7 + 10 â€“ 15 = 2

â‡’ 2 = 2

Therefore, Eulerâ€™s formula is verified.

(ii)No. of Faces (F) = 9No. of Vertices (V) = 9

No. of Edges (E) = 16

By Using Eulerâ€™s formula: F + V – E = 2 and Substituting the values, we get

â‡’ 9 + 9 – 16 = 2

â‡’ 2 = 2

Therefore, Eulerâ€™s formula is verified.

**Question 7:** Using Eulerâ€™s formula find the unknown.

Faces |
? |
5 |
20 |

Vertices |
6 |
? |
12 |

Edges |
12 |
9 |
? |

**Solution:**

(i)No. of Faces (F) = F

No. of Vertices (V) = 6

No. of Edges (E) = 12

By Using Eulerâ€™s formula: F + V â€“ E = 2 and Substituting the values, we get

â‡’ F + 6 â€“ 12 = 2

â‡’ F = 2 + 6

â‡’ F = 8

Therefore, No. of Faces (F) = 8

(ii)No. of Faces (F) = 5

No. of Vertices (V) = V

No. of Edges (E) = 9

By Using Eulerâ€™s formula: F + V â€“ E = 2 and Substituting the values, we get

â‡’ 5 + V â€“ 9 = 2

â‡’ E = 2 + 4

â‡’ E = 6

Therefore, No. of Vertices (V) = 6

(iii)No. of Faces (F) = 20

No. of Vertices (V) = 12

No. of Edges (E) = E

By Using Eulerâ€™s formula: F + V â€“ E = 2 and Substituting the values, we get

â‡’ 20 + 12 â€“ E = 2

â‡’ E = 32 – 2

â‡’ E = 30

Therefore, No. of Edges (E) = 30

**Question 8:** Can a polyhedron have 10 faces, 20 edges and 15 vertices?

**Solution:**

Since every polyhedron satisfies Eulerâ€™s formula, therefore checking if polyhedron can have 10 faces, 20 edges and 15 vertices.

No. of Faces (F) = 10

No. of Vertices (V) = 15

No. of Edges (E) = 20

By Using Eulerâ€™s formula: F + V â€“ E = 2 and Substituting the values, we get

â‡’ 10 + 15 â€“ 20 = 2

â‡’ -5 = 2

As Eulerâ€™s formula is not satisfied, therefore polyhedron cannot have 10 faces, 20 edges and 15 vertices.

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