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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.

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Question

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.

Answer in Brief
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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  .

 

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Chapter 18: Surface Areas and Volume of a Cuboid and Cube - Exercise 18.3 [Page 35]

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R.D. Sharma Mathematics [English] Class 9
Chapter 18 Surface Areas and Volume of a Cuboid and Cube
Exercise 18.3 | Q 2 | Page 35

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