GSEB Class 8 Maths Solutions Chapter 10 Visualizing Solid Shapes Exercise 10.3

Question 1. Can a polyhedron 3. a square and four triangles?
Answer: A polyhedron is enclosed by four or more polygonal faces.
1. No, it is not possible that a polyhedron has 3 triangles for its faces.
2. Yes, 4 triangles can be the faces of a polyhedron.
3. Yes, a square and 4 triangles can be the faces of a polyhedron.
In simple words: A polyhedron must have at least four flat sides, which are polygons. You cannot make a polyhedron with only three triangles. However, you can make one with four triangles (like a pyramid) or one with a square base and four triangular sides (another type of pyramid).

Exam Tip: Remember that a polyhedron must enclose a space, which requires a minimum number of faces to form a closed figure. Euler's formula is a good way to check the validity of polyhedra.

 

Question 2. Is it possible to have a polyhedron with any given number of faces? Hint: Think of a pyramid.
Answer: Yes, it can be possible only if the number of faces is four or more than four.
In simple words: A polyhedron must always have four or more faces. It cannot have fewer than four faces.

Exam Tip: The simplest polyhedron is a tetrahedron, which has 4 faces, 4 vertices, and 6 edges. Any fewer faces will not form a closed 3D shape.

 

Question 3. Which are prisms among the following?
(i) A nail
(ii) Unsharpened pencil
(iii) A table weight
(iv) A box
Answer: Since, a prism is a polyhedron having two of its faces congruent and parallel, whereas other faces are parallelograms.
1. No, a nail is not a prism.
2. Yes, an unsharpened pencil is a prism.
3. No, a table weight is not a prism.
4. Yes, a box is a prism.
In simple words: A prism is a 3D shape with two identical and parallel ends, and its other sides are parallelograms. A nail and a table weight do not fit this description, but an unsharpened pencil and a box do.

Exam Tip: To identify a prism, look for two bases that are the same shape and size, and are parallel to each other. The connecting faces should be parallelograms (or rectangles in a right prism).

 

Question 4.
1. How are prisms and cylinders alike?
2. How are pyramids and cones alike?
Answer:
1. Both prisms and cylinders have their base and top as congruent faces and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base gets larger and larger.
2. The pyramid and cones are alike because their lateral faces meet at a vertex. Also, a pyramid becomes a cone as the number of sides of its base gets larger and larger.
In simple words: Prisms and cylinders both have matching, parallel top and bottom faces. A prism turns into a cylinder if its base has many, many sides. Similarly, pyramids and cones both come to a point at the top. A pyramid becomes a cone if its base has many, many sides.

Exam Tip: The key similarity lies in how their bases and side faces behave: parallel and congruent for prisms/cylinders, and tapering to a point for pyramids/cones.

 

Question 5. Is a square prism same as a cube? Explain?
Answer: No, not always, because it can be a cuboid also.
In simple words: A square prism is not always a cube. It can also be a cuboid, which has rectangular faces, not necessarily square ones. A cube is a special kind of square prism where all its sides are equal in length.

Exam Tip: A cube is a specific type of cuboid where all faces are squares. A square prism has square bases but its height can vary, making its lateral faces rectangles instead of squares.

 

Question 6. Verify Euler's formula for these solids?
Answer:
(i) In figure (i), we have F = 7, V = 10, and E = 15.
\( F + V = 7 + 10 = 17 \)
\( F + V - E = 17 - 15 = 2 \)
i.e., \( F + V - E = 2 \)
Thus, Euler's formula is verified.
(ii) In figure (ii), we have F = 9, V = 9, and E = 16.
\( F + V = 9 + 9 = 18 \)
\( F + V - E = 18 - 16 = 2 \)
i.e., \( F + V - E = 2 \)
Thus, Euler's formula is verified.
In simple words: For both shapes, if you add the number of faces and vertices, then subtract the number of edges, you always get 2. This shows Euler's formula works for these solids.

Exam Tip: Carefully count the faces (F), vertices (V), and edges (E) of each solid. For Euler's formula \( F + V - E = 2 \) to be true, the shape must be a convex polyhedron.

 

Question 7. Using Euler's formula find the unknown?

Answer:
1. For (i): Here, \( V = 6 \) and \( E = 12 \).
Since \( F + V - E = 2 \)
\( \implies F + 6 - 12 = 2 \)
\( \implies F - 6 = 2 \)
\( \implies F = 2 + 6 \)
\( \implies F = 8 \)
2. For (ii): Here, \( F = 5 \) and \( E = 9 \).
Since \( F + V - E = 2 \)
\( \implies 5 + V - 9 = 2 \)
\( \implies V - 4 = 2 \)
\( \implies V = 2 + 4 \)
\( \implies V = 6 \)
3. For (iii): Here, \( F = 20 \) and \( V = 12 \).
Since \( F + V - E = 2 \)
\( \implies 20 + 12 - E = 2 \)
\( \implies 32 - E = 2 \)
\( \implies E = 32 - 2 \)
\( \implies E = 30 \)
In simple words: We used Euler's formula \( F + V - E = 2 \) for each case. For the first, we found the number of faces to be 8. For the second, we calculated the number of vertices as 6. For the third, we determined the number of edges to be 30.

Exam Tip: Remember Euler's formula \( F + V - E = 2 \). This formula is important for polyhedra, linking the number of faces, vertices, and edges. Make sure to apply the algebraic steps correctly to solve for the unknown value.

 

Question 8. Can polyhedron have 10 faces, 20 edges and 15 vertices?
Answer: Here, \( F = 10, E = 20 \), and \( V = 15 \).
We have: \( F + V - E = 2 \)
\( \implies 10 + 15 - 20 = 2 \)
\( \implies 25 - 20 = 2 \)
\( \implies 5 = 2 \) which is not true.
i.e., \( F + V - E \neq 2 \)
Thus, such a polyhedron is not possible.
In simple words: If we use Euler's formula with the given numbers (10 faces, 15 vertices, 20 edges), the calculation does not equal 2. This means a polyhedron with those specific counts of faces, vertices, and edges cannot exist.

Exam Tip: Always verify Euler's formula \( F + V - E = 2 \) when asked if a polyhedron with specific properties is possible. If the formula does not hold, then such a polyhedron cannot exist.