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\( 2\) cubes each of volume \(64\;\;{cm}^{3}\)are joined end to end. Find the surface area of the resulting cuboid.
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\( 2\) cubes each of volume \(64\;\;{cm}^{3}\)are joined end to end. Find the surface area of the resulting cuboid.

Answer

The surface area of resulting cuboid is \(160\ cm^2\)
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Solution

Answer: 
The diagram is given as:
Solution for 2 cubes each of volume 64;;{cm}^{3}are joined end to end. Find the surface area of the resulting cuboid.
Given, 
The Volume (\(V\)) of each cube is\(=64\;{cm}^{3}\) 
 \(\because Volume\;={\left. \left(side\right)\right. }^{3}\)
Let a be the side of the given cube , so,
 \({a}^{3}=64\)
\(\Rightarrow a=4\;cm\)
Now the resulting cuboid will be
Solution for 2 cubes each of volume 64;;{cm}^{3}are joined end to end. Find the surface area of the resulting cuboid.
The length of cuboid (l)\(=a+a=4+4=8\;cm\)
breadth(b)\(=a=4\;cm\)
height(h)\(=a=4\;cm\)
So, Surface Area of Cuboid \(= 2(lb + bh + lh)\) ;  where l is length ,b is breadth and h is height of the cuboid.
\(=2(8\times 4+4\times 4+4\times 8)\;{cm}^{2}\) 
\(=2(32+16+32)\;{cm}^{2}\) 
\(=(2\times 80)\;{cm}^{2}=160\;{cm}^{2}\)
Thus , surface area of cuboid is \(160\;{cm}^{2}\).
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