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If the length of the largest straight rod that can be put inside a cub

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If the length of the largest straight rod that can be put inside a cub [#permalink]

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New post 20 Aug 2017, 06:27
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Question Stats:

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If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the surface area of
the cuboid cannot be more than

A 100 m2
B 200 m2
C 400 m2
D 600 m2
E Cannot be determined
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If the length of the largest straight rod that can be put inside a cub [#permalink]

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New post 20 Aug 2017, 06:41
vishwash wrote:
If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the surface area of
the cuboid cannot be more than

A 100 m2
B 200 m2
C 400 m2
D 600 m2
E Cannot be determined


hi..

cuboid should be cube here to get the longest diagonal
let the side of cuboid be a..
the straightest rod can have longest length equal to the diagonal which will be \(\sqrt{3a^2}=10....a^2 = 100/3\)
surface area = \(6*a^2=6*\frac{100}{3} = 200\)

B
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If the length of the largest straight rod that can be put inside a cub [#permalink]

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New post 20 Aug 2017, 08:26
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vishwash wrote:
If the length of the largest straight rod that can be put inside a cuboid is 10 m, then the surface area of
the cuboid cannot be more than

A 100 m2
B 200 m2
C 400 m2
D 600 m2
E Cannot be determined

The rod is equivalent to the longest straight line in a cuboid -- a "space" diagonal that runs between two vertices that are not on the same side (e.g. upper right corner on front side to lower left corner on back side).

Assume the cuboid is a cube; all cubes are cuboids, though not vice versa. For a fixed surface area, a cube yields the greatest volume for a cuboid, and hence the longest diagonal.* Alternatively, given just one measure (the space diagonal), the only way to calculate surface area of this cuboid is if s side equals length, width, and height -- namely, if it is a cube (instead of a rectangular box).

If you know the length of that diagonal, you can find a cube's side length because sides are equal.

Formula for an interior diagonal of a polyhedron is a variation on Pythagorean theorem:

\(l^2 + w^2 + h^2 = d^2\)

l = w = h = side s, and d = diagonal, given as 10

\(s^2 + s^2 + s^2 = 10^2\)
3\(s^2\) = 100
\(s^2\) = \(\frac{100}{3}\)

Don't calculate \(s\). Surface area uses \(s^2\).

Surface area of cube: \(6s^2\)

6 * \(\frac{100}{3}\) = \(200 m^2\)

Answer B

*among other reasons, because the sum of squares of two or more numbers, where sum is constant (e.g. the given length of the space diagonal), is greatest when the squared numbers are equal
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If the length of the largest straight rod that can be put inside a cub   [#permalink] 20 Aug 2017, 08:26
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