5 Marks Questions

Question 15 Marks

Using Euler's formula find the unknown:

Faces	$?$	$5$	$20$
Vertices	$6$	$?$	$12$
Edges	$12$	$9$	$?$

Answer

We know that the Euler's formula is: $F + V = E + 2$
$i.$ The number of vertices $V$ is $6$ and the number of edges $E$ is $12$.
Using Euler's formula:
$F + 6 = 12 + 2$
$F + 6 = 14$
$F = 14 - 6$
$F = 8$
So, the number of faces in this polyhedron is 8.
$ii.$ Faces, $F = 5$
Edges, $E = 9$
We have to find the number of vertices.
Putting these values in Euler's formula:
$5 + V = 9+ 25 + V = 11$
$V = 11 - 5$
$V = 6$
So, the number of vertices in this polyhedron is $6$.
$iii.$ Number of faces $F = 20$
Number of vertices $V = 12$
Using Euler's formula:
$20 + 12 = E + 2$
$32 = E + 2$
$E + 2 = 32$
$E = 32 - 2$
$E = 30$
So, the number of edges in this polyhedron is $30$.

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Question 25 Marks

Match the following figures:

Answer

$i.$ The given figure is a cuboid with sides $4, 4$ and $6$ units.
Area of a rectangle $=$ length $\times $ width
$\therefore$ Area of the rectangular face sith sides $4$.

$ii.$ The given figure is a cuboid with sides $3, 3$ and $8$.Area of a rectangle $=$ length $\times$ width
$\therefore$ Area of the rectangular face sith sides $3$ and $3 = 3 \times 3 = 9$ And the area of the other face with sides $3$ and $8 = 3 \times 8 = 24$
Thus, the net for given figure will have four faces with area $24$ and two faces with area $9$.
Observe that net $(i)$ satisfies this.
Thus, the net of figure $(b)$ is net $(i)$.

$iii.$ The given figure is a cuboid with sides $3, 4$ and $6$.
Area of a rectangle = length $\times $ width
$\therefore$ Area of the rectangular face sith sides $3$ and $4 = 3 \times 4 = 12$,
Area of the rectangular face with sides $4$ and $6 = 4 \times 6 = 24$
And, area of the other face with sides $3$ and $6 = 3 \times 6 = 18$
Thus, the net for given figure will have two faces with area $24$, two faces wit area $18$ and two faces with area $12$.
Observe that net $(ii)$ satisfies this.
Thus, the net of figure $(c)$ is net $(ii)$.

$iv.$ The given figure is a cuboid with sides $3, 3$ and $9$.
Area of a rectangle $=$ length $\times$ width
$\therefore$ Area of the rectangular face with sides $3$ and $3 = 3 \times 3 = 9$, And
area of the other face with sides $3$ and $9 = 3 \times 9 = 27$
Thus, the net for given figure will have four faces with area $27$ and two faces with area $9$.
Observe that net $(iii)$ satisfies this.
Thus, the net of figure $(d)$ is net $(iii)$.

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