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(D) The sum of the surface areas of the three smaller cubes is given by $6 	imes (3^2 + 4^2 + 5^2) = 6 	imes (9 + 16 + 25) = 6 	imes 50 = 300 \, cm

Vedclass · Accepted Answer

$25:18$

Vedclass · Answer

(D) The sum of the surface areas of the three smaller cubes is given by $6 	imes (3^2 + 4^2 + 5^2) = 6 	imes (9 + 16 + 25) = 6 	imes 50 = 300 \, cm^2$.The volume of the large cube formed by melting these is the sum of their volumes: $V = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 \, cm^3$.Let the side of the large cube be $a$. Then $a^3 = 216$,which implies $a = 6 \, cm$.The surface area of the large cube is $6 	imes a^2 = 6 	imes 6^2 = 6 	imes 36 = 216 \, cm^2$.The ratio of the total surface area of the smaller cubes to the surface area of the large cube is $300 : 216$.Dividing both by their greatest common divisor,$12$,we get $300/12 : 216/12 = 25 : 18$.

$A$ 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?

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