# How much metal is needed to cast a cubical metal box with outer side 10 inches long if the thickness of its walls should be exactly 2 inches?

It would take

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To find the amount of metal needed to cast a cubical metal box with outer side length 10 inches and wall thickness of 2 inches, we first need to find the dimensions of the inner cube.

The inner side length of the cube is the outer side length minus twice the thickness of the walls: [ \text{Inner side length} = \text{Outer side length} - 2 \times \text{Wall thickness} ]

[ \text{Inner side length} = 10 - 2 \times 2 = 10 - 4 = 6 \text{ inches} ]

Now, we calculate the surface area of the inner cube. Since all sides of a cube are equal, we use the formula: [ \text{Surface area of cube} = 6 \times \text{side length}^2 ]

[ \text{Surface area of cube} = 6 \times 6^2 = 6 \times 36 = 216 \text{ square inches} ]

The total surface area of the metal box includes the inner cube's surface area and the surface area of the walls. Since the box has six walls (one for each face of the inner cube), each with the same dimensions as the inner cube but with added thickness, we calculate the surface area of one wall and multiply it by six: [ \text{Surface area of one wall} = \text{Inner side length} \times \text{Wall thickness} ] [ \text{Surface area of one wall} = 6 \times 2 = 12 \text{ square inches} ]

Now, we calculate the total surface area of the metal box: [ \text{Total surface area of box} = \text{Surface area of inner cube} + 6 \times \text{Surface area of one wall} ] [ \text{Total surface area of box} = 216 + 6 \times 12 = 216 + 72 = 288 \text{ square inches} ]

Therefore, 288 square inches of metal are needed to cast the cubical metal box.

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