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 cubes and the large cube?<br /><br />
A) 2 : 1 B) 3 : 2 C) 25 : 18 D) 27 : 20​

Answer 1

Answer:

261

Step-by-step explanation:

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 cubes and the large cube? Explanation: Volume of the large cube = (33 + 43 + 53) = 216 cm3.

Answer 2

Answer:

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

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

${x}^{3} = {3}^{3} + {4}^{3} + {6}^{3} \\ = > x {}^{3} = 27 + 64 + 125 = 216 \\ = > x = \sqrt[3]{216} = 6 \: cm$

Therefore, required ratio = Total surface are of smaller cubes ÷ Surface area of smaller cubes

$= 6( {3}^{2} ) + 6( {4}^{2} ) + 6( {5}^{2} ) \div 6(6 {}^{2} ) \\ = \frac{6 \times 9 + 6(4 {}^{2}) + 6(5 {}^{2}) }{6(6) {}^{2} }$

$= \frac{6 \times 9 + 6 \times 16 + 6 \times 25}{6 \times 36} \\ = \frac{54 + 96 + 150}{216} \\ = \frac{300}{216} \\ = \frac{300}{216} = \frac{25}{18}$

= 25 : 18.