In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 17 Graphs and Polyhedra for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 17 Graphs and Polyhedra pdf, free KSEEB solutions for Class 10 Maths Chapter 17 Graphs and Polyhedra book pdf download. Now you will get step by step solution to each question.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra (English Medium)
KSEEB SSLC Solutions for Class 10 Maths Chapter 17 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.
KSEEB SSLC Solutions for Class 10 Maths Chapter 17 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 |
KSEEB SSLC Solutions for Class 10 Maths Chapter 17 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.
KSEEB SSLC Solutions for Class 10 Maths Chapter 17 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.
All Chapter KSEEB Solutions For Class 10 Maths
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All Subject KSEEB Solutions For Class 10
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