Answer

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Hint: To calculate the length of all the edges, multiply the length of each edge with the number of edges. To calculate the cost of painting the outside of the box on all sides, calculate the total surface area of the cube, which is \[6{{a}^{2}}\]. Use a unitary method to calculate the total cost of painting. To calculate the volume of liquid which the box can hold, calculate the volume of the cube using the formula \[{{a}^{3}}\], where ‘a’ is the length of the edge of the cube.

Complete step-by-step answer:

We have a cube whose length of edge is 5 m. We have to calculate the total length of edges, cost of painting the outside of the box and volume of liquid that the box can hold.

(a) We have to calculate the total length of edges. We know that a cube has 12 edges.

To find the total length of edges, we will multiply the length of each edge with the number of edges of the cube.

Thus, the total length of all the edges \[=12\times 5=60m\].

(b) We have to calculate the total cost of painting all the outside surfaces of the cube, which are painted at the rate of \[Rs.5\] per m sq.

We will firstly calculate the total surface area of the cube.

We know that the total surface area of the cube whose edge is ‘a’ is given by \[6{{a}^{2}}\].

Substituting \[a=5m\], the total surface area of cube \[=6{{a}^{2}}=6{{\left( 5 \right)}^{2}}=150{{m}^{2}}\].

We will now calculate the total cost of painting the surfaces of the cube. We know that \[1{{m}^{2}}\] of area is painted for \[Rs.5\].

To calculate the cost of painting \[x{{m}^{2}}\] of area, we will multiply x by 5.

Thus, the cost of painting \[150{{m}^{2}}\]\[=150\times 5=Rs.750\].

Hence, the cost of painting all the outer surfaces of the cube is \[Rs.750\].

(c) We have to calculate the volume of the cube.

We know that the volume of the cube of side ‘a’ is given by \[{{a}^{3}}\].

Substituting \[a=5m\], the volume of cube \[={{5}^{3}}=125{{m}^{3}}\].

Hence, the volume of cube is \[125{{m}^{3}}\].

Note: A cube is a three dimensional object whose all sides are of equal length. A cube has 12 edges, 8 vertices and 6 faces. It’s necessary to keep the units in mind while finding out the values of different measurements of the cube. The total surface area has measurements in \[{{m}^{2}}\], while the volume has measurements in \[{{m}^{3}}\].

Complete step-by-step answer:

We have a cube whose length of edge is 5 m. We have to calculate the total length of edges, cost of painting the outside of the box and volume of liquid that the box can hold.

(a) We have to calculate the total length of edges. We know that a cube has 12 edges.

To find the total length of edges, we will multiply the length of each edge with the number of edges of the cube.

Thus, the total length of all the edges \[=12\times 5=60m\].

(b) We have to calculate the total cost of painting all the outside surfaces of the cube, which are painted at the rate of \[Rs.5\] per m sq.

We will firstly calculate the total surface area of the cube.

We know that the total surface area of the cube whose edge is ‘a’ is given by \[6{{a}^{2}}\].

Substituting \[a=5m\], the total surface area of cube \[=6{{a}^{2}}=6{{\left( 5 \right)}^{2}}=150{{m}^{2}}\].

We will now calculate the total cost of painting the surfaces of the cube. We know that \[1{{m}^{2}}\] of area is painted for \[Rs.5\].

To calculate the cost of painting \[x{{m}^{2}}\] of area, we will multiply x by 5.

Thus, the cost of painting \[150{{m}^{2}}\]\[=150\times 5=Rs.750\].

Hence, the cost of painting all the outer surfaces of the cube is \[Rs.750\].

(c) We have to calculate the volume of the cube.

We know that the volume of the cube of side ‘a’ is given by \[{{a}^{3}}\].

Substituting \[a=5m\], the volume of cube \[={{5}^{3}}=125{{m}^{3}}\].

Hence, the volume of cube is \[125{{m}^{3}}\].

Note: A cube is a three dimensional object whose all sides are of equal length. A cube has 12 edges, 8 vertices and 6 faces. It’s necessary to keep the units in mind while finding out the values of different measurements of the cube. The total surface area has measurements in \[{{m}^{2}}\], while the volume has measurements in \[{{m}^{3}}\].

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