Class 8 SELINA Solutions Maths Chapter 19: Representing 3-D in 2-D
Representing 3-D in 2-D Exercise Ex. 19
Solution 1(i)
Correct option: (b) 4
A tetrahedron has 4 faces.
Solution 1(ii)
Correct option: (b) 8
An octahedron has 8 faces.
Solution 1(iii)
Correct option: (d) 5 vertices and 5 faces
A rectangular pyramid has 5 vertices and 5 faces.
Solution 1(iv)
Correct option: (c) F + V – E = 2
Euler’s formula is F + V – E = 2
Solution 1(v)
Correct option: (c) 30
Number of faces, F = 12
Number of vertices, V = 20
Then, by Euler’s formula,
F + V – E = 2
⇒ 12 + 20 – E = 2
⇒ E = 30
Solution 2
Number of faces, F = 7
Number of vertices, V = 10
Number of edges = E
Then, by Euler’s formula,
F + V – E = 2
⇒ 7 + 10 – E = 2
⇒ E = 15
Solution 3
(i) Pentagonal pyramid
Number of faces = 6
Number of vertices = 6
Number of edges = 10
(ii) Hexagonal prism
Number of faces = 8
Number of vertices = 12
Number of edges = 18
Solution 4(i)
For a given figure,
Number of faces, F = 5
Number of vertices, V = 5
Number of edges, E = 8
Now, F + V – E = 5 + 5 – 8 = 2
Hence, verified.
Solution 4(ii)
For a given figure,
Number of faces, F = 9
Number of vertices, V = 9
Number of edges, E = 16
Now, F + V – E = 9 + 9 – 16 = 2
Hence, verified.
Solution 4(iii)
For a given figure,
Number of faces, F = 7
Number of vertices, V = 10
Number of edges, E = 15
Now, F + V – E = 7 + 10 – 15 = 2
Hence, verified.
Solution 5
Number of faces, F = 8
Number of vertices, V = 16
Number of edges, E = 26
Now, F + V – E = 8 + 16 – 26 = –2 ≠ 2
Therefore, a polyhedron cannot have 8 faces, 26 edges and 16 vertices.
Solution 6
(i) No, a polyhedron is a three-dimensional figure with at least 4 polygonal faces.
(ii) Yes, a polyhedron can have 4 triangles only. A tetrahedron has 4 triangular faces.
(iii) Yes, a polyhedron can have a square and four triangles. A square pyramid has a square and four triangular faces.
Solution 7
Euler’s formula: F + V – E = 2
(i) F = x, V = 15, E = 20
x + 15 – 20 = 2
x – 5 = 2
x = 7
(ii) F = 6, V = y, E = 8
6 + y – 8 = 2
y – 2 = 2
y = 4
(iii) F = 14, V = 26, E = z
14 + 26 – z = 2
40 – z = 2
z = 38
Solution 8
The least number of planes that can enclose a solid is 4.
The name of the solid is tetrahedron.
Solution 9
Yes, a square prism is same as a cube.
Solution 10
Two nets of cuboid with dimensions 6 cm × 4 cm × 2 cm are as follows:
Solution 11
The sum of opposite faces of dice is 7.
Here,
Opposite face of a will be 5.
⇒ a + 5 = 7
⇒ a = 2
Opposite face of b will be 6.
⇒ b + 6 = 7
⇒ b = 1
Opposite face of c will be is 4.
⇒ c + 4 = 7
⇒ c = 3
Solution 12
Nets for Hexagonal prism:
Nets for Pentagonal pyramid:
Representing 3-D in 2-D Exercise TEST YOURSELF
Solution 1(i)
Correct option: (a) 12
By Euler’s formula, F + V – E = 2
⇒ 20 + V – 30 = 2
⇒ V – 10 = 2
⇒ V = 12
Solution 1(ii)
Correct option: (b) triangular prism
It is a net of triangular prism.
Solution 1(iii)
Correct option: (c) 15
A pentagonal prism has 15 edges. Hence, Joseph will need 15 straws.
Solution 1(iv)
Correct option: (c) 14
A hexagonal pyramid has 12 edges,
By Euler’s formula, F + V – E = 2
Therefore, F + V = 2 + E = 2 + 12 = 14
Solution 1(v)
Correct option: (b) 4
A triangular pyramid has 4 faces.
Solution 2
A net is prepared for three-dimensional solids. Since rectangle is a two-dimensional figure, it does not have a net. It can be drawn as follows:
Solution 3
Number of faces, F = 15
Number of vertices, V = 20
Number of edges, E = 30
Now, F + V – E = 15 + 20 – 30 = 5 ≠ 2
Therefore, a polyhedron cannot have 15 faces, 30 edges and 20 vertices.
Solution 4
The net of a square pyramid is as follows:
Solution 5
A hexagonal pyramid has 12 edges.
Solution 6
The two-dimensional representation of a triangular prism is as follows:
Solution 7
Euler’s formula: F + V – E = 2
For a given polyhedron, F = 10, V = 8, E = ?
10 + 8 – E = 2
18 – E = 2
E = 16
Hence, the number of edges is 16.
Solution 8
(i) Triangular prism
(ii) Triangular prism
(iii) Hexagonal pyramid