Question

Question!
$\: \: \: \:$
Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 12√3 cm. Find the edges of the three cubes​

Answer 1

Answer:

$Let \: the \: edge \: of \: 3 \: cube \: 3x \: 4x \: and \: 5x$

$Volume \: of \: 3 \: cubes \\ V1 = (3x ^{2} ) = 27x ^{2} \\ V2 = ( {4x}^{2} ) = 64x ^{2} \\ V3 = ( {5x}^{2} ) = {125x}^{2}$

$Total \: volume = V = V1 + V2 + V3 = 27x ^{2} + {64x}^{2} + {125x}^{2}$

$V = 216x {}^{3}$

$Volume \: of \: new \: cube = Volume \: of \: cubes = V$

$Let \: the \: side \: of \: new \: cube = 12 \sqrt{3cm}$

$\sqrt{a ^{2} + a ^{2} + a ^{2}} = 12 \sqrt{3cm}$

$a \sqrt{3} = 12 \sqrt{3cm}$

$a = 12cm = 6x(from \: i)$

$x= 2cm$

$The \: \: edges \: \: of \: \: 3 \: cubes \: \: are$

$a1 = 3x = 3 \times 2 = 6cm \\ a2 = 4x = 4 \times 2 = 8cm \\a3 = 5x = 5 \times 2 = 10cm$

I hope this might help you with ☺

Answer 2

Answer:

Given :-

• Three cubes of a metal whose edges are in the ratio of 3 : 4 : 5 are melted and converted into a single cube whose diagonal is 12√3 cm.

To Find :-

• What is the edges of the three cubes.

Solution :-

Let,

$\mapsto \bf Edges\: of\: 1^{st}\: Cube =\: 3x$

$\mapsto \bf Edges\: of\: 2^{nd}\: Cube =\: 4x$

$\mapsto \bf Edges\: of\: 3^{rd}\: Cube =\: 5x$

As we know that :-

$\bigstar \: \: \sf\boxed{\bold{\pink{Volume_{(Cube)} =\: a^3}}}\: \: \: \bigstar$

According to the question :

$\footnotesize \implies \bf Volume\: of\: three\: cubes =\: Volume\: of\: new\: cube$

$\implies \sf (3x)^3 + (4x)^3 + (5x)^3 =\: a^3$

$\implies \sf 27x + 64x + 125x =\: a^3$

$\implies \sf 216x^3 =\: a^3$

$\implies \sf 6x =\: a$

$\implies \sf\bold{\blue{a =\: 6x}}$

Now,

$\bigstar$ Diagonal of a new single cube formed is 12√3 cm.

As we know that :

$\bigstar \: \: \sf\boxed{\bold{\pink{Diagonal_{(Cube)} =\: \sqrt{3}a}}}\: \: \: \bigstar$

So, according to the question by using the formula we get,

$\implies \bf {\cancel{\sqrt{3}}}a =\: 12{\cancel{\sqrt{3}}}$

$\implies \sf a =\: 12$

By putting a = 6x we get,

$\implies \sf 6x =\: 12$

$\implies \sf x =\: \dfrac{\cancel{12}}{\cancel{6}}$

$\implies \sf\bold{\purple{x =\: 2}}$

Hence, the required edges of the three cubes are :

$\footnotesize \dashrightarrow \sf\bold{\red{Edge\: of\: 1^{st}\: Cube =\: 3x =\: 3 \times 2 =\: 6\: cm}}$

$\footnotesize \dashrightarrow \sf\bold{\red{Edge\: of\: 2^{nd}\: Cube =\: 4x =\: 4 \times 2 =\: 8\: cm}}$

$\footnotesize \dashrightarrow \sf\bold{\red{Edge\: of\: 3^{rd}\: Cube =\: 5x =\: 5 \times 2 =\: 10\: cm}}$

$\therefore$ The edges of the three cubes are 6 cm , 8 cm and 10 cm .