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If the area of 3 adjecent faces of a rectangular block are

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Director
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If the area of 3 adjecent faces of a rectangular block are [#permalink]

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New post 08 Jan 2005, 12:42
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If the area of 3 adjecent faces of a rectangular block are in the ratio 2:3:4 and its vol is 9000. Then what is the length of the shorter side .
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shortest side [#permalink]

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New post 08 Jan 2005, 13:35
Lets assume that the sides are x, y, and z:

Then xy:xz:zy are in (2:3:4) (Given)

xy/xz=y/z=2/3
xy/yz=x/z=2/4
xz/yz=x/y=3/4

xyz=9000

y=2z/3 and x=3y/4=z/2

xyz = (z/2)*(2z/3)*(z)=9000

z=30, y=20, x=15:

Thus shortest side is 15
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New post 08 Jan 2005, 13:40
x = 15

x is the smallest side.

xy/ xz/yz = 2/3/4

xy/xz = 2/3

3/2 y = z

x/y = 3/4 ----1

x = 3/4 y ---2

xyz = 9000----3



8/3 x^3 = 9000 = 27000/ 8 = 30/2


x = 15
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 [#permalink]

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New post 08 Jan 2005, 13:58
I did it this way

let the sides be a b c

then given is

ab: bc : ca = 2: 3: 4 --------- I

volume = abc = 9000

divide I by 9000 or abc

ab /abc : bc /abc : ca /abc = 1/c: 1/a : 1/b = 2 / 9000 : 3 /9000 : 4 /9000

c:a:b: = 450:300:225 = 30 : 20 : 15

Shortest side = 15

but we make sure abc = 9000 which it is 30*20*15
  [#permalink] 08 Jan 2005, 13:58
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If the area of 3 adjecent faces of a rectangular block are

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