KSEEB SSLC Solutions for Class 10 Maths Chapter 17 Graphs and Polyhedra

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.

 NRA
(i)6___10
(ii)53____
(iii)____49
(iv)____712
(v)15___20

Solution :

Euler’s formula: N + R = A + 2

 Nodes(N)Regions(R)Arcs(A)
(i)66 + R = 10 + 2   6 + R = 12 R = 610
(ii)535 + 3 = A + 2   8 = A + 2 A = 6
(iii)N + 4 = 9 + 2   N + 4 = 11 N = 749
(iv)N + 7 = 12 + 2   N + 7 = 14 N = 7712
(v)1515 + R = 20 + 2   15 + R = 22 R = 720


Question 2:

Draw the graphs for the given values of N, A and R.

 NRA
(i)7510
(ii)8612
(iii)547
(iv)558


Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.1-2s1
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.1-2s2

Question 3:

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.1-3

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.

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:

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-1 .

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-1s
NodeOrderType
K5Odd
L5Odd
M5Odd
N5Odd
O5Odd

Question 2:

Find the order and type of each node in the following graph:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-2s
NodeOrderType
P4Even
Q3Odd
R3Odd

Question 3:

Find the order and type of each node in the following graph:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-3s
NodeOrderType
A4Even
B4Even
C4Even
D2Even

Question 4:

Find the order and type of each node in the following graph:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-4

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-4s
NodeOrderType
E3Odd
F6Even
G3Odd
H4Even

Question 5:

Find the order and type of each node in the following graph:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-5

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-5s
NodeOrderType
A8Even
B3Odd
C3Odd

Question 6:

Find the order and type of each node in the following graph:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-6

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.2-6
NodeOrderType
A2Even
B2Even
C2Even
D2Even
E4Even
F4Even
G4Even
H4Even
P3Odd
Q3Odd

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-1

Solution :

NodeOrderType
A3Odd
B3Odd
C3Odd
D3Odd
E3Odd
F3Odd
G3Odd
H3Odd

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-2

Solution :

NodeOrderType
P2Even
Q3Odd
R3Odd
S2Even
T4Even

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-3

Solution :

NodeOrderType
K2Even
L2Even
M6Even
N2Even
O2Even
P2Even
Q2Even

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-4

Solution :

NodeOrderType
D2Even
E6Even
F2Even
G2Even
H2Even

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-5

Solution :

NodeOrderType
D2Even
E5Odd
F2Even
G2Even
H2Even
I1Odd

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-6

Solution :

NodeOrderType
A2Even
B6Even
C6Even
D2Even
E2Even
F2Even

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-7

Solution :

NodeOrderType
K1Odd
L4Even
M3Odd
N4Even
O1Odd
P1Odd
Q3Odd
R1Odd

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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.3-8

Solution :

NodeOrderType
A6Even
B6Even
C6Even
D6Even
E1Odd
F1Odd
G1Odd
H1Odd

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.

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-1

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-1
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.
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-2

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-2
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.

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-3

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-3
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.

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-4

Solution :

KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-4
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:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-IIa s
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:
KSEEB SSLC Solutions for Class 10 Maths – Graphs and Polyhedra Ex-17.5-IIb s
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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