Question

A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. what is the ratio of the total surface areas of the smaller cube to the large cube?​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

Let x be the edge of the large cube. Then,

$x {}^{3} + 3 {}^{3} + {4}^{3} + {5}^{3} \\ = > x {}^{3} = 27 + 64 + 125 \\ = 216$

$= > x = \sqrt[3]{216} = 6 \: cm$

Therefore required ratio

= Total surface area of smaller cubes ÷ Surface area of larger cube

$= \frac{6(3) {}^{2} + 6(4) {}^{2} + 6 ({5})^{2} }{ 6(6) {}^{2} }$

$= \frac{6 \times 9 + 6 \times 16 + 6 \times 25}{6(6) {}^{2} }$

$= \frac{54 + 96 + 150}{216} \\ \\ = \frac{300}{216} \\ \\ \\ = \frac{25}{18}$

= 25 : 18.

Answer 2

Answer:

Volume of the large cube = (3³ + 4³ + 5³) = 216 cm3

216=6³

Step-by-step explanation:

So the side length of the large cube is 6 cm

Since there are 6 faces of a cube so the surface area of the small cubes taken together

Sm =6(9 +16 +25) = 6(50) = 300 cm2

The surface area of the large cube

So = 6 ×36 = 216 cm2

The ratio Sm/Sl =300/216 = 25/18

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