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# Three Cubes of a Metal Whose Edges Are in the Ratios 3 : 4 : 5 Are Melted and Converted into a Single Cube Whose Diagonal is - Mathematics

Answer in Brief

Three cubes  of a metal whose edges are in the ratios 3 : 4 : 5 are melted and converted into a single cube whose diagonal is  $12\sqrt{3}$. Find the edges of the three cubes.

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#### Solution

The three cubes of metal are in the ratio 3 : 4 : 5.
Let the edges of the cubes be 3x, 4x and 5x.
Volume of the three cubes will be

$V_1 = \left( 3x \right)^3$

$V_2 = \left( 4x \right)^3$

$V_3 = \left( 5x \right)^3$

Diagonal of the single cube = $12\sqrt{3} cm$

We know diagonal of the cube =  $a\sqrt{3} = 12\sqrt{3}$

Hence, the side of the cube = 12 cm
Volume of the bigger cube $V_b = \left( 12 \right)^3$

Volume of the three cubes = Volume of the single

$\left( 3x \right)^3 + \left( 4x \right)^3 + \left( 5x \right)^3 = \left( 12 \right)^3$

$\Rightarrow 27 x^3 + 64 x^3 + 125 x^3 = 1728$

$\Rightarrow 216 x^3 = 1728$

$\Rightarrow x^3 = \frac{1728}{216} = 8$

$\Rightarrow x = 2$

Hence, the edges of the three cubes  will be  $3 \times \left( 2 \right), 4 \times \left( 2 \right), 5 \times \left( 2 \right) = 6, 8, 10$ cm.

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 14 Surface Areas and Volumes
Exercise 14.1 | Q 12 | Page 28
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