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Difficult Geometry Quesiton

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Manager
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Difficult Geometry Quesiton [#permalink]

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New post 04 Aug 2009, 20:23
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Manager
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Re: Difficult Geometry Quesiton [#permalink]

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New post 05 Aug 2009, 03:02
for small triangle
Area = 1/2 (s*h) .........h is the height

for bigger triangle
Area = 1/2 (S*H) .........H is the height

From the question we have
1/2 (S*H) = 2 * [1/2 (s*h)]

so S/s = h/H....equation A

For similar triangles

h/H = s/S

so h = (H*s)/S ....equation B

From A and B we get


S= s *sqrt( 2 )


HTH

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Re: Difficult Geometry Quesiton [#permalink]

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New post 05 Aug 2009, 04:50
in your calculation where does the extra '2' go??

From the question we have
1/2 (S*H) = 2 * [1/2 (s*h)]

so S/s = h/H....equation A

it shud be>> S/s= 2 h/H

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Re: Difficult Geometry Quesiton [#permalink]

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New post 05 Aug 2009, 04:54
apoorvasrivastva wrote:
From the question we have
1/2 (S*H) = 2 * [1/2 (s*h)]

so S/s = h/H....equation A


Hey apoorvasrivastva, the answer is good, I think you made a typo in eq A...
S/s = 2 h/H

Anyway, we can also do this by applying a property of similar triangles, which is:
For similar triangles, the ratio of the corresponding areas is equal to the ratio of the squares of any of the corresponding lengths.

therefore, Triangle Area(right)/Triangle Area(left) = 2/1 = S^2/s^2 => S = s sqrt(2)
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Re: Difficult Geometry Quesiton [#permalink]

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New post 05 Aug 2009, 07:45
scarish wrote:
apoorvasrivastva wrote:
From the question we have
1/2 (S*H) = 2 * [1/2 (s*h)]

so S/s = h/H....equation A


Hey apoorvasrivastva, the answer is good, I think you made a typo in eq A...
S/s = 2 h/H

Anyway, we can also do this by applying a property of similar triangles, which is:
For similar triangles, the ratio of the corresponding areas is equal to the ratio of the squares of any of the corresponding lengths.

therefore, Triangle Area(right)/Triangle Area(left) = 2/1 = S^2/s^2 => S = s sqrt(2)



hey u r right..i made a typo..sorry folks!!

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Senior Manager
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Re: Difficult Geometry Quesiton [#permalink]

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New post 05 Aug 2009, 13:05
if two circles/spheres or whatever of the same shape...

D and d is the length of respective sides
S and s is the (surface) area
V and v is the volume of a 3d shape...

then usually S/s = (D/d)^2, V/v = (D/d)^3...i think this is usually the case. so here would be 2/1 = (D/d)^2, D/d = sqrt(2)

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Re: Difficult Geometry Quesiton   [#permalink] 05 Aug 2009, 13:05
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