Question
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 $(iii)$.
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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$6$
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$1$
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Expenditure
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Family $A$
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Family $B$
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Food
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Clothing
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