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 1](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRB3tP_ZeOuF6vog46uTajLdRRdvLlzyAoVKnvJHpZDsHSg8yY2eG0xQQm5jmSVRpmxwGPrsc2xw6kvA9eXxB3zjA-4C4zK3tfrDxmCUZ_LSErIqmXvcjxdhTDavkg201TJpqChMYSMb1XD9g8yUq8hKQE7nmw16lxc0uRcqRFJZVMfuuRyv3YPmn9wQ/s16000/1.png)
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?
![Question 3](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLULwDq1AE2bPfbVJ4Jauuz8AgKYIw269k6aCasYDha-qpPyHghSASFhF2HrlQHgTTWUftAKrlEvaJhxmRPlewGsBIvt0JKJ87d_URK2ie0IUkiJBZ6HTyW5V1Lb2VLhYb2B70Cfik3li4QxGumbiWtS01ksmjlZsBwy4uPQ26VUIGkmISWGFwMVd0BQ/s16000/2.png)
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.
![Question 6](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyYWG2JKG5LJlglU-PU-56l_Lih8iK5-dBiFET3Je_UN6vNLnqllIgiJ2vi9oLz6v07YC2YEnOy7jSfwGyz4WLELYgii1UjGPKXrnPSq8Jf1AdgKmL9AMSyWWnhWPNbamhdjh3PCsWqitTgOWi3RgJBM8m6nVSAka9ttpR7hjdo_s10mdmY6V3BQM3HA/s16000/3.png)
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.
![Question 7](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhSbFzBovQt2mBvM8eVj1aEBWo4si6HeKVWjUcz1PUSin2fhXDMW4ryHPABW0g7plwlTWSiu14PpEDsEai6s8BDRVIYPjdJHVGSIHIPJQ18IleapkGtlgBTTPMvehucxN45p_k-WwT_bNEcaEnYzvUQyiFChckkwrAVZm8EdGdYc9sFMLlCgl2mIToD8Q/s16000/4.png)
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.
No comments:
Post a Comment