Question

2 cubes each of volume 64 $c{m}^3$ are joined end to end. Then the surface area of resulting cuboid (in $c{m}^3$)

• A. 160
• B. 96
• C. 64
• D. 4

Answer 1

The correct option is A 160

The side of each cube = $\sqrt[3]{364}$ = 4 cm.
Also, surface area of a cuboid = $2\left(\right(l\times b)+(b\times h)+(l\times h\left)\right)$
Here, l = 8 cm, b = h = 4 cm
So, surface area of cuboid = $2\left(\right(8\times 4)+(4\times 4)+(8\times 4\left)\right)$
= 2(32 + 16 + 32)
= 160 $c{m}^2$
Also, after joining two cubes, the surface area of resulting cuboid is $10a^2$, where 'a' is the side of original cube.