# Three Cubes of Metal Whose Edges Are in the Ratio 3 : 4 : 5 Are Melted Down in to a Single Cube Whose Diagonal is 12 √ 3 Cm. Find the Edges of Three Cubes. - Mathematics

Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are melted down in to a single cube whose diagonal is 12 sqrt(3) cm. Find the edges of three cubes.

#### Solution

The edges of the three cubes are in the ratio 3 : 4 : 5.

So, let the edges be 3x cm, 4x cm, 5x cm.

The diagonal of new cube is 12sqrt(3) cm

We need to find the edges of three cubes

Here, volume of the resulting cube,

V = (3x)^3 + (4x)^3 + (5x)^3

=27x^3 + 64x^3 + 125x^3

= 216x^3

Let,

l → Edge of the resulting cube

So, diagonal of the cube= sqrt(3l), so

12sqrt(3) = sqrt(3l)

Hence,

l = 12 cm

Now;

V=1^3

216x^3 = 12^3

(6x)^3 = 12^3

x = 2

The edges of the three cubes are,

3x = 3× 2

= 6cm

4x = 4× 2

= 8 cm

5x = 5 × 2

= 10 cm

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

Concept: Surface Area of a Cuboid
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#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 18 Surface Areas and Volume of a Cuboid and Cube
Exercise 18.3 | Q 2 | Page 35

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