KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra

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 :

1. 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.
2. 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.
3. 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.
4.  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.
5. N = 4, R = 8, A = 10
   N + R = A + 2
   4 + 8 = 10 + 2
   12 = 12
   ∴ L.H.S. = R.H.S.
6. ∴ 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.
7. 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.
8. 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.