## 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?

Answer:

i) No, a polyhedron can’t have 3 triangles for its faces because polyhedron must have edges meeting at the vertices called points.

ii) Yes, a polyhedron can have 4 triangles for its faces because all the edges are meeting at the vertices called points.

iii) Yes, a polyhedron can have a square and four triangles for its faces because all the edges are meeting at the vertices called points.

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

Answer: Yes, it is possible to have a polyhedron with the number of faces is greater than or equal to 4. Example: Pyramid with 4 faces.

Question 3: Which are prisms among the following?

Answer: ii) Unsharpened pencil and iii) A table weight are prisms.

Question 4:

i) How are prisms and cylinders alike?

ii) How are pyramids and cones alike?

Answer:

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.

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

Answer: No, every square prism is not same as cube. It may be a cuboid also.

Question 6: Verify Euler’s formula for these solids.

Answer:

i)

Number of faces = 7

Number of edges = 15

Number of vertices = 10

by using Euler’s formula

= F + V - E = 2

= 7 + 10 - 15 = 2

= 17 - 15 = 2

= 2 = 2

Hence, verified.

ii)

Number of faces = 9

Number of edges = 16

Number of vertices = 9

by using Euler’s formula

= F + V - E = 2

= 9 + 9 - 16 = 2

= 18 - 16 = 2

= 2 = 2

Hence, verified.

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

Answer:

i)

Number of faces = x

Number of vertices = 6

Number of edges = 12

by using Euler’s formula

= F + V - E = 2

= x + 6 - 12 = 2

= x - 6 = 2

= x = 6 + 2

= x = 8

Therefore, the number of the faces of the solid is 8.

ii)

Number of faces = 5

Number of vertices = x

Number of edges = 9

by using Euler’s formula

= F + V - E = 2

= 5 + x - 9 = 2

= 5 + x = 2 + 9

= 5 + x = 11

= x = 11 - 5

= x = 6

Therefore, the number of the vertices of the solid is 6.

iii)

Number of faces = 20

Number of vertices = 12

Number of edges = x

by using Euler’s formula

= F + V - E = 2

= 20 + 12 - x = 2

= 32 - x = 2

= 32 - 2 = x

= 30 = x

Therefore, the number of the edges of the solid is 30.

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

Answer:

Number of faces = 10

Number of vertices = 15

Number of edges = 20

by using Euler’s formula

= F + V - E = 2

= 10 + 15 - 20 = 2

= 25 - 20 = 2

= 5 ≠ x

Therefore, a polyhedron can’t have 10 faces, 20 edges and 15 vertices.

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