Question

# $2$cubes each of volume$64{\mathrm{cm}}^{3}$ are joined end to end. Find the surface area of the resulting cuboid.

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Solution

## Step 1:Find the side of cube. $\begin{array}{rcl}\mathrm{Volume}\mathrm{of}\mathrm{cube}& =& 64{\mathrm{cm}}^{3}={\left(\mathrm{Side}\right)}^{3}\\ {\left(\mathrm{Side}\right)}^{3}& =& 64\\ \mathrm{Side}& =& {\left(64{\mathrm{cm}}^{3}\right)}^{\frac{1}{3}}\\ & =& {\left(2×2×2×2×2×2\right)}^{\frac{1}{3}}\\ & =& 2×2\\ \therefore \mathrm{Side}& =& 4\mathrm{cm}\end{array}$Step 2:Find the surface area of the resulting cuboid. Upon joining the two cubes the dimension of the cuboid will be, Length$=8\mathrm{cm}$, breadth$=4\mathrm{cm}$, Height$=4\mathrm{cm}$$\begin{array}{rcl}\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{cuboid}& =& 2\left(\mathrm{lb}+\mathrm{bh}+\mathrm{hl}\right)\\ & =& 2\left(8×4+4×4+4×8\right)\\ & =& 2\left(32+16+32\right)\\ & =& 2\left(80\right)\\ & =& 160{\mathrm{cm}}^{2}\end{array}$Hence, the surface area of the resulting cuboid is$160{\mathrm{cm}}^{2}$.

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