# RBSE Solutions for Class 8 Maths Chapter 8 Visualization of Solids Additional Questions

RBSE Solutions for Class 8 Maths Chapter 8 Visualization of Solids Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 8 Visualization of Solids Additional Questions.

 Board RBSE Textbook SIERT, Rajasthan Class Class 8 Subject Maths Chapter Chapter 8 Chapter Name Visualization of Solids Exercise Additional Questions Number of Questions 37 Category RBSE Solutions

## Rajasthan Board RBSE Class 8 Maths Chapter 8 Visualization of Solids Additional Questions

I. Objective Type Questions

Question 1.
Plane figures are called
(a) multi-dimensional
(b) three-dimensional
(c) two-dimnsional
(d) none of the above

Question 2.
Number of vertices in a cuboid are
(a) 12
(b) 8
(c) 6
(d) 4

Question 3.
Every solid shape is made(RBSESolutions.com)up of various two dimensional figures. These are called
(a) edges
(b) vertices
(c) faces
(d) cuboid

Question 4.
Euler’s formula is
(a) F + V = E + 2
(b) F – V = E – 2
(c) V + E = F + 2
(d) V – E = F – 2

Question 5.
The point where the edges of a solid, is called
(a) face
(b) surface
(c) vertex
(d) pyramid

Question 6.
An example of three dimensional figure is
(a) cuboid
(b) circle
(c) square
(d) rectangle

Question 7.
Number of faces in a cuboid are
(a) 6
(b) 8
(c) 12
(d) 10

Question 8.
Number of edges in a cube are
(a) 8
(b) 12
(c) 6
(d) 14

Question 9.
Number of faces in a cube
(a) 12
(b) 8
(c) 6
(d) 10

1. (c)
2. (b)
3. (c)
4. (a)
5. (c)
6. (a)
7. (a)
8. (b)
9. (c).

II. Fill in the blanks

Question 1.
Cuboid is a___shape.

Question 2.
3D objects will have___views from different angles.

Question 3.
For any polyhedron, F + V =___ is true.

Question 4.
There is no reference or perspective in a ____

Question 5
___is very important(RBSESolutions.com)for drawing a picture but it is not relevant for a map.

1. three-dimensional
2. different
3. E + 2
4. map
5. Perspective.

III. Very Short Answer Type Questions

Question 1.
Write number of edges and faces in a triangular prism.
Number of edges in a triangular prism = 9 and number of faces in a triangular prism = 5.

Question 2.
What are two dimensional shapes?
The plane figures(RBSESolutions.com)having two measurements, length and breadth, are called two dimensional shapes.

Question 3.
Give 3 examples of two dimensional shapes.
Triangle, rectangle circle etc.

Question 4.
Give 3 examples of three dimensional shapes.
Cuboid, sphere, cylinder etc.

Question 5.
What is the definition of Prism?
A prism is a polyhedron(RBSESolutions.com)whose base and top are congruent polygons and whose other faces, i.e., lateral faces are parallelogram in shapes.

Question 6.
What do you mean by pyramid?
A polyhedron having a polygonal base and triangular sides with a common vertex, is called a pyramid.

Question 7.
Find the number of faces in polyhedron having vertices 10 and edges 16.
Solution
Number of vertices (V) = 10, Number of edges (E) =16, Number of faces (F) = ?
Euler formula V + F = E + 2
⇒ 10 + F = 16 + 2
or F = 16 + 2 – 10 = 8

Question 8.
Define a regular polyhedrons.
A polyhedron is said to be(RBSESolutions.com)regular if its faces are made up of regular polygons and the same number of faces meet at each vertex.

Question 9.
Is it possible to have a polyhedron with any given number of faces?
Yes, it is possible only if the number of faces are greatest then or equal to four.

Question 10.
Is a square, prism same as a cube? Explain.
Yes, It can be a cube. But it can be a cuboid also.

Question 11.
Can a polyhedron have(RBSESolutions.com)10 faces, 20 edges and 15 vertices?
Since, F + V = E + 2
As 10 + 15 ≠ 20 + 2
∴ A polyhedron cannot have 10 faces, 20 edges and 15 vertices.

Question 1.
One vertex of a cube in cut equidistant from her three sides as shown in fig. How many faces and vertices in new fig.

Solution:
Number of faces (F) = 7
Number of vertices (V) = 10

Question 2.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution
We know that polyhedron is a solid, which is bounded by four or more polygonal faces in such a way that pairs of faces meet along edges and three or more edges meet in each vertex, therefore,
(i) It is not possible(RBSESolutions.com)that a polyhedron has 3 triangles for its faces.
(ii) 4 triangles can be the faces of a polyhedron.
(iii) A square and 4 triangles can be the faces of a polyhedron.

Question 3.
Which are prisms among the following?

Solution
We know that a prism is a polyhedron whose base and top are congruent polygons and lateral faces are parallelogram. Therefore,
(i) A nail is not a prism.
(ii) An unsharpened pencil is a prism.
(iii) A table(RBSESolutions.com)weight is not a prism.
(iv) A box is a prism.

Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution
(i) A prism becomes a cylinder provided the number of sides of its base becomes large and larger.
(ii) A pyramid becomes a(RBSESolutions.com)cone provided the number of sides of its base becomes larger and larger.

Question 5.
Using Euler’s formula find the missing numbers.

 Polyhedron Face Vertex Edges A ? 6 12 B 5 ? 9

Solution
For A Polyhedron
Number of faces = ?;
Number of vertices = 6,
Number of Edges = 12
By using Euler’s formula
F + V = E + 2
or F = E + 2 – V = 12 + 2 – 6
F = 14 – 6 = 8
∴ Number of faces F = 8
For B Polyhedron
Number of faces = 5;
Number of vertices = ?,
Number of Edges = 9 ,
By using Euler’s formula
F + V = E + 2
or V = E + 2 – F = 9 + 2 – 5 = 6
∴ Number of vertices = 6

Question 6.
Find number of edges in a polyhedron which have 9 vertices and 9 faces.
Solution
Number of vertices (V) = 9
Number(RBSESolutions.com)of faces (F) = 9
To Find : Number of edges.
Euler’s formula : V + F = E + 2
Then, 9 + 9 = E + 2
or 9 + 9 – 2 = E
Hence, E = 16

Question 7.
In a polyhedron, number of faces is 5 and number(RBSESolutions.com)of edges is 9. Find the number of vertices.
Solution
Number of faces (F) = 5
Number of Edges (E) = 9
Number of Vertices (V) = ?
We know by Euler’s formula
F + V = E + 2
5 + V = 9 + 2
5 + V = 11
V = 11 – 5
V = 6
∴Number of vertices = 6

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