# An open box of length 1.5 m, breadth 1 m and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust-proof paint. At a rate of 150 rupees per sqm, how much will it cost to paint the box?

Answer

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Hint: First of all, get the metal sheet required by finding the total surface area of the open box that is S = 2 (bh + lh) + lb and then multiply twice of this area with cost to paint per \[{{\text{m}}^{2}}\] to get the total cost of the paint required.

Complete step-by-step answer:

We are given an open box whose length, breadth and height are 1.5 m, 1 m and 1 m respectively. We have to find the metal sheet required to make this box. Also, the inside and outside surface area of the box is painted at a rate of 150 rupees/\[{{\text{m}}^{2}}\]. We have to find the cost to paint the box.

First all, we have to find the sheet required to make this box. To get the total sheet required, we have to find the total surface area of the given box.

Let us consider the surface area of the box to be S.

We know that this box is in the shape of a cuboid. So, the total surface area of cuboid = 2 (lb + bh + hl) where l, b and h are the length, breadth and height of the cuboid.

But as we are given that this box is an open box that is it is open at the top. So, we must subtract the area of the upper face from the total surface area of the cuboid to get the required surface area of the given box. So, we get

Surface area of box (S) = 2 (lb + bh + hl) – (Area of the upper face)

We know that area of the upper face of box = l x b

So, we get, S = 2(lb + bh + hl) – lb

Therefore, we get surface area of the box (S) = 2 (bh + hl) + lb …..(1)

Now, we know that the metal sheet required = Surface area of the box

So now, we substitute the length of the box = 1.5 m, breadth of box = 1 m and height of the box = 1 m in equation (1) to get the metal sheet required. So, we get,

S = Metal sheet required \[=2\left[ \left( 1\times 1+1\times 1.5 \right)+\left( 1\times 1.5 \right) \right]\text{ }{{\text{m}}^{2}}\]

\[S=\left[ 2\left( 1+1.5 \right)+1.5 \right]\text{ }{{\text{m}}^{2}}\]

\[S=6.5\text{ }{{\text{m}}^{2}}\]

So, we get the metal sheet required to make the given box equal to 6.5 sq.m.

Now, we are given that the inside and outside surface area of the box is painted. So, we get,

The total area of the box to be painted = Inside the surface area of the box + outside surface area of the box

As we know that the inside surface area = outside surface area = S for this box, so we get,

Total area of the box to be painted = S + S = 2S.

By substituting the value of \[S=6.5{{\text{m}}^{2}}\], we get,

The total area of the box to be painted = \[2\times 6.5{{\text{m}}^{2}}=13{{\text{m}}^{2}}\].

We are given that cost to paint the box = \[\text{Rs}\text{.150/}{{\text{m}}^{2}}\]

So, we get the cost to paint the outside and inside surface area of the box \[=\text{Rs}\left( 150\times 13 \right)=\text{Rs}.1950\]

Therefore, we get the total cost to paint the box equal to Rs.1950.

Note: Here, many students forget to subtract the area of the upper face of the cuboid box. So, this must be kept in mind whenever the box is open. Also, students must note that we have to paint both inside and outside of the box. So, we have to add the surface area twice to get the area to be painted. So, students should keep this point in mind.

Complete step-by-step answer:

We are given an open box whose length, breadth and height are 1.5 m, 1 m and 1 m respectively. We have to find the metal sheet required to make this box. Also, the inside and outside surface area of the box is painted at a rate of 150 rupees/\[{{\text{m}}^{2}}\]. We have to find the cost to paint the box.

First all, we have to find the sheet required to make this box. To get the total sheet required, we have to find the total surface area of the given box.

Let us consider the surface area of the box to be S.

We know that this box is in the shape of a cuboid. So, the total surface area of cuboid = 2 (lb + bh + hl) where l, b and h are the length, breadth and height of the cuboid.

But as we are given that this box is an open box that is it is open at the top. So, we must subtract the area of the upper face from the total surface area of the cuboid to get the required surface area of the given box. So, we get

Surface area of box (S) = 2 (lb + bh + hl) – (Area of the upper face)

We know that area of the upper face of box = l x b

So, we get, S = 2(lb + bh + hl) – lb

Therefore, we get surface area of the box (S) = 2 (bh + hl) + lb …..(1)

Now, we know that the metal sheet required = Surface area of the box

So now, we substitute the length of the box = 1.5 m, breadth of box = 1 m and height of the box = 1 m in equation (1) to get the metal sheet required. So, we get,

S = Metal sheet required \[=2\left[ \left( 1\times 1+1\times 1.5 \right)+\left( 1\times 1.5 \right) \right]\text{ }{{\text{m}}^{2}}\]

\[S=\left[ 2\left( 1+1.5 \right)+1.5 \right]\text{ }{{\text{m}}^{2}}\]

\[S=6.5\text{ }{{\text{m}}^{2}}\]

So, we get the metal sheet required to make the given box equal to 6.5 sq.m.

Now, we are given that the inside and outside surface area of the box is painted. So, we get,

The total area of the box to be painted = Inside the surface area of the box + outside surface area of the box

As we know that the inside surface area = outside surface area = S for this box, so we get,

Total area of the box to be painted = S + S = 2S.

By substituting the value of \[S=6.5{{\text{m}}^{2}}\], we get,

The total area of the box to be painted = \[2\times 6.5{{\text{m}}^{2}}=13{{\text{m}}^{2}}\].

We are given that cost to paint the box = \[\text{Rs}\text{.150/}{{\text{m}}^{2}}\]

So, we get the cost to paint the outside and inside surface area of the box \[=\text{Rs}\left( 150\times 13 \right)=\text{Rs}.1950\]

Therefore, we get the total cost to paint the box equal to Rs.1950.

Note: Here, many students forget to subtract the area of the upper face of the cuboid box. So, this must be kept in mind whenever the box is open. Also, students must note that we have to paint both inside and outside of the box. So, we have to add the surface area twice to get the area to be painted. So, students should keep this point in mind.

Last updated date: 20th Sep 2023

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