Question

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is __________.

Answer 1

Let the edge of three small metallic cubes be 3x, 4x and 5x, respectively.

Suppose the edge of the big cube is a cm.

It is given that the three small metallic cubes are melted to form a big cube.

∴ Volume of big cube = Sum of the volumes of three small cubes

⇒ a³ = (3x)³ + (4x)³ + (5x)³

⇒ a³ = 27x³ + 64x³ + 125x³ = 216x³

⇒ a³ = (6x)³

⇒ a = 6x .....(1)

It is given that the length of diagonal to the big cube is 18 cm.

$\therefore \sqrt{3}a=18\mathrm{cm}$ (Length of the diagonal of the cube = $\sqrt{3}$ × Side)

$\Rightarrow 6\sqrt{3}x=18$ [Using (1)]

$\Rightarrow x=\frac{18}{6\sqrt{3}}=\sqrt{3}\mathrm{cm}$

Now,

Edge of the smallest cube = 3x = $3\sqrt{3}$ cm

Surface area of the cube = 6 × (Side)²

∴ Total surface area of the smallest cube = $6\times {\left(3\sqrt{3}\right)}^2$ = 6 × 27 = 162 cm²

Thus, the the total surface area of the smallest cube is 162 cm².

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is $\underline{162{\mathrm{cm}}^2}$.