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

Given:

Two cubes, each of volume $64\ cm^3$ are joined end to end.

To do:

We have to find the surface area of the resulting cuboid.

Solution:

Volume of each cube $= 64\ cm^3$

This implies,

Edge of the cube $= \sqrt[3]{64}$

$=4\ cm$

The length of the cuboid formed by joining the cubes $(l) = 4 + 4$

$= 8\ cm$

Breadth of the cuboid $(b) = 4\ cm$

Height of the cuboid $(h) = 4\ cm$

Therefore,

Surface area of the resulting cuboid $= 2(lb + bh + lh)$

$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$

$= 2(32 + 16 + 32)$

$= 2 \times 80$

$= 160\ cm^2$ .

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