KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra (English Medium)
Exercise 17.1:
Question 1:
Using Euler’s formula, complete the following table.
N | R | A | |
(i) | 6 | ___ | 10 |
(ii) | 5 | 3 | ____ |
(iii) | ____ | 4 | 9 |
(iv) | ____ | 7 | 12 |
(v) | 15 | ___ | 20 |
Solution :
Euler’s formula: N + R = A + 2
Nodes(N) | Regions(R) | Arcs(A) | |
(i) | 6 | 6 + R = 10 + 2
6 + R = 12 R = 6 |
10 |
(ii) | 5 | 3 | 5 + 3 = A + 2
8 = A + 2 A = 6 |
(iii) | N + 4 = 9 + 2
N + 4 = 11 N = 7 |
4 | 9 |
(iv) | N + 7 = 12 + 2
N + 7 = 14 N = 7 |
7 | 12 |
(v) | 15 | 15 + R = 20 + 2
15 + R = 22 R = 7 |
20 |
Question 2:
Draw the graphs for the given values of N, A and R.
N | R | A | |
(i) | 7 | 5 | 10 |
(ii) | 8 | 6 | 12 |
(iii) | 5 | 4 | 7 |
(iv) | 5 | 5 | 8 |
Solution :
Question 3:
Solution :
- N = 4, R = 4, A = 6
N + R = A + 2
4 + 4 = 6 + 2
8 = 8
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 3, R = 5, A = 6
N + R = A + 2
3 + 5 = 6 + 2
8 = 8
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 4, R = 4, A = 6
N + R = A + 2
4 + 4 = 6 + 2
8 = 8
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 7, R = 5, A = 10
N + R = A + 2
7 + 5 = 10 + 2
12 = 12
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 4, R = 8, A = 10
N + R = A + 2
4 + 8 = 10 + 2
12 = 12
∴ L.H.S. = R.H.S. - ∴ The graph satisfies Euler’s formula.
N = 12, R = 8, A = 18
N + R = A + 2
12 + 8 = 18 + 2
20 = 20
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 5, R = 7, A = 10
N + R = A + 2
5 + 7 = 10 + 2
12 = 12
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula. - N = 9, R = 14, A = 21
N + R = A + 2
9 + 14 = 21 + 2
23 = 23
∴ L.H.S. = R.H.S.
∴ The graph satisfies Euler’s formula.
Exercise 17.2:
Question 1:
Find the order and type of each node in the following graph:
.
Solution :
Node | Order | Type |
K | 5 | Odd |
L | 5 | Odd |
M | 5 | Odd |
N | 5 | Odd |
O | 5 | Odd |
Question 2:
Find the order and type of each node in the following graph:
Solution :
Node | Order | Type |
P | 4 | Even |
Q | 3 | Odd |
R | 3 | Odd |
Question 3:
Find the order and type of each node in the following graph:
Solution :
Node | Order | Type |
A | 4 | Even |
B | 4 | Even |
C | 4 | Even |
D | 2 | Even |
Question 4:
Find the order and type of each node in the following graph:
Solution :
Node | Order | Type |
E | 3 | Odd |
F | 6 | Even |
G | 3 | Odd |
H | 4 | Even |
Question 5:
Find the order and type of each node in the following graph:
Solution :
Node | Order | Type |
A | 8 | Even |
B | 3 | Odd |
C | 3 | Odd |
Question 6:
Find the order and type of each node in the following graph:
Solution :
Node | Order | Type |
A | 2 | Even |
B | 2 | Even |
C | 2 | Even |
D | 2 | Even |
E | 4 | Even |
F | 4 | Even |
G | 4 | Even |
H | 4 | Even |
P | 3 | Odd |
Q | 3 | Odd |
Exercise 17.3:
Question 1:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
A | 3 | Odd |
B | 3 | Odd |
C | 3 | Odd |
D | 3 | Odd |
E | 3 | Odd |
F | 3 | Odd |
G | 3 | Odd |
H | 3 | Odd |
Number of even nodes: 0
Number of odd nodes: 8
Since the network contains more than two odd nodes and no even nodes, it is not a transversable network.
Question 2:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
P | 2 | Even |
Q | 3 | Odd |
R | 3 | Odd |
S | 2 | Even |
T | 4 | Even |
Number of even nodes: 3
Number of odd nodes: 2
Since the network contains only two odd nodes, it is a transversable network.
Question 3:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
K | 2 | Even |
L | 2 | Even |
M | 6 | Even |
N | 2 | Even |
O | 2 | Even |
P | 2 | Even |
Q | 2 | Even |
Number of even nodes: 7
Number of odd nodes: 0
Since the network contains all even nodes, it is a transversable network.
Question 4:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
D | 2 | Even |
E | 6 | Even |
F | 2 | Even |
G | 2 | Even |
H | 2 | Even |
Number of even nodes: 5
Number of odd nodes: 0
Since the network contains all even nodes, it is a transversable network.
Question 5:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
D | 2 | Even |
E | 5 | Odd |
F | 2 | Even |
G | 2 | Even |
H | 2 | Even |
I | 1 | Odd |
Number of even nodes: 4
Number of odd nodes: 2
Since the network contains only two odd nodes, it is a transversable network.
Question 6:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
A | 2 | Even |
B | 6 | Even |
C | 6 | Even |
D | 2 | Even |
E | 2 | Even |
F | 2 | Even |
Number of even nodes: 6
Number of odd nodes: 0
Since the network contains all even nodes, it is a transversable network.
Question 7:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
K | 1 | Odd |
L | 4 | Even |
M | 3 | Odd |
N | 4 | Even |
O | 1 | Odd |
P | 1 | Odd |
Q | 3 | Odd |
R | 1 | Odd |
Number of even nodes: 2
Number of odd nodes: 6
Since the network contains more than two odd nodes, it is not a transversable network.
Question 8:
Verify the transversablity of the following network and state the reason.
Solution :
Node | Order | Type |
A | 6 | Even |
B | 6 | Even |
C | 6 | Even |
D | 6 | Even |
E | 1 | Odd |
F | 1 | Odd |
G | 1 | Odd |
H | 1 | Odd |
Number of even nodes: 4
Number of odd nodes: 4
Since the network contains more than two odd nodes, it is not a transversable network.
Exercise 17.5:
Question I(1):
Write the number of faces, edges and vertices for the following solid figure and verify Euler’s formula.
Solution :
Number of faces, F = 6
Number of vertices, V = 8
Number of edges, E = 12
Verification of Euler’s formula:
F + V = E + 2
6 + 8 = 12 + 2
14 = 14
∴ L.H.S. = R.H.S.
Question I(2):
Write the number of faces, edges and vertices for the following solid figure and verify Euler’s formula.
Solution :
Number of faces, F = 8
Number of vertices, V = 12
Number of edges, E = 18
Verification of Euler’s formula:
F + V = E + 2
8 + 12 = 18 + 2
20 = 20
∴ L.H.S. = R.H.S.
Question I(3):
Write the number of faces, edges and vertices for the following solid figure and verify Euler’s formula.
Solution :
Number of faces, F = 7
Number of vertices, V = 7
Number of edges, E = 12
Verification of Euler’s formula:
F + V = E + 2
7 + 7 = 12 + 2
14 = 14
∴ L.H.S. = R.H.S.
Question I(4):
Write the number of faces, edges and vertices for the following solid figure and verify Euler’s formula.
Solution :
Number of faces, F = 6
Number of vertices, V = 6
Number of edges, E = 10
Verification of Euler’s formula:
F + V = E + 2
6 + 6 = 10 + 2
12 = 12
∴ L.H.S. = R.H.S.
Question II(a):
Verify Euler’s formula for Dodecahedron.
Solution :
Dodecahedron:
Number of faces, F = 12
Number of vertices, V = 20
Number of edges, E = 30
Verification of Euler’s formula:
F + V = E + 2
12 + 20 = 30 + 2
32 = 32
∴ L.H.S. = R.H.S.
Question II(b):
Verify Euler’s formula for Icosahedron.
Solution :
Icosahedron:
Number of faces, F = 20
Number of vertices, V = 12
Number of edges, E = 30
Verification of Euler’s formula:
F + V = E + 2
20 + 12 = 30 + 2
32 = 32
∴ L.H.S. = R.H.S.