Question

2 cubes each of volume \[64\,c{{m}^{3}}\] are joined end to end. Find the surface area of the resulting cuboid.

(a) \[128\,c{{m}^{3}}\]
(b) \[130\,c{{m}^{3}}\]
(c) \[160\,c{{m}^{3}}\]
(d) none of these

Answer 1

Hint: We will first draw the figure and then everything will get clear. From the drawn figure we will see that the length of the cuboid is 2 times the edge of the cube and all other dimensions will be the same. We know that the total surface area of a cuboid is \[2\times (lb+bh+hl)\].

Complete step-by-step answer:

Drawing the figure from the above inputs mentioned in the question, we get,

Now it is mentioned that the volume of 1 cube is \[64\,c{{m}^{3}}\]. We know that \[V={{a}^{3}}.......(1)\].

So now substituting the given volume in equation (1) and solving for a we get,

\[\Rightarrow 64={{a}^{3}}......(2)\]

Now we can write 64 as 4 to the power cube in equation (2) and rearranging we get,

\[\begin{align}

  & \Rightarrow {{a}^{3}}={{4}^{3}} \\

 & \Rightarrow a=4......(3) \\

\end{align}\]

Now before finding the surface area of the resulting cuboid let us find all the dimensions of the cuboid.

So from the figure we can see that the length of the cuboid is a+a. So using this information we get,

\[l=a+a......(4)\]

Now substituting the value of a from equation (3) in equation (4) we get,

\[l=4+4=8\,cm\]

And breadth and height in the cuboid remains same as the edge in the cube that is a.

Total surface area of a cuboidal box \[=2\times (lb+bh+hl)..........(5)\]

Putting the value of h, l and b in equation (5) we get,

\[\Rightarrow 2\times \left[ (8\times 4)+(4\times 4)+(4\times 8) \right]=2\times 80=160\,c{{m}^{2}}\]

Hence the surface area of the cuboid is \[160\,c{{m}^{2}}\]. So the correct answer is option (c).

Note: Remembering the formula is the key here and chances of mistakes are when we substitute value of b or h in l and value of b or l in h. And as we are calculating area and unit of area is \[c{{m}^{2}}\] so we should not forget to write the unit otherwise marks will be deducted.