3 cubes of metal whose edges are in the ratio | vedclass

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Question

(A) Let the edges of the three cubes be $3k, 4k,$ and $5k$ meters respectively.The volume of a cube is given by $(	ext{edge})^3$. Therefore,the volum

Vedclass · Accepted Answer

$24 	ext{ m}, 32 	ext{ m}, 40 	ext{ m}$

Vedclass · Answer

(A) Let the edges of the three cubes be $3k, 4k,$ and $5k$ meters respectively.The volume of a cube is given by $(	ext{edge})^3$. Therefore,the volumes of the three cubes are $(3k)^3 = 27k^3$,$(4k)^3 = 64k^3$,and $(5k)^3 = 125k^3$.The total volume of the single cube formed by melting these three is the sum of their individual volumes:$V = 27k^3 + 64k^3 + 125k^3 = 216k^3$.If the edge of the new cube is $a$,then $a^3 = 216k^3$,which implies $a = \sqrt[3]{216k^3} = 6k$.The diagonal of a cube with edge $a$ is given by $a\sqrt{3}$.Given that the diagonal of the new cube is $48\sqrt{3} 	ext{ m}$,we have:$a\sqrt{3} = 48\sqrt{3} \implies a = 48 	ext{ m}$.Since $a = 6k$,we have $6k = 48$,which gives $k = 8$.Thus,the edges of the three cubes are:$3k = 3 	imes 8 = 24 	ext{ m}$$4k = 4 	imes 8 = 32 	ext{ m}$$5k = 5 	imes 8 = 40 	ext{ m}$.

$3$ cubes of metal whose edges are in the ratio $3: 4: 5$ are melted into a single cube,the length of whose diagonal is $48 \sqrt{3} \text{ m}$. Calculate the edges of the three cubes.

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