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Given two similiar triangles, if the area of one triangle, [#permalink]
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10 Nov 2005, 07:45
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Given two similiar triangles, if the area of one triangle, side s, is twice the area of the second triangleside S, then in terms of s, S =
(sqrt2/2)s
(sqrt3/2)s
sqrt2s
sqrt3s
2s
Pleae show your calcs
A



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the stem sounds kinda awkward.. could you clarify a little? are the commas where they should be?
I calculated something and came up with ssqrt(1/2)
still not sure because of the stem.



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area 1 = s^2/2....area 2 = S^2/2....s^2/2 = 2*S^2/2...s * sqrt(1/2) = S...its not given...
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can someone tell me why area is s^2/2 ?
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bewakoof wrote: can someone tell me why area is s^2/2 ?
i assumed a triangle with 1:1:sqrt(2)...area of this triangle with side s is s*s*1/2=s^2/2
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christoph wrote: bewakoof wrote: can someone tell me why area is s^2/2 ? i assumed a triangle with 1:1:sqrt(2)...area of this triangle with side s is s*s*1/2=s^2/2
i see so you assumed it was a right triangle. what if it is not?? coz it doesnt say..
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bewakoof wrote: christoph wrote: bewakoof wrote: can someone tell me why area is s^2/2 ? i assumed a triangle with 1:1:sqrt(2)...area of this triangle with side s is s*s*1/2=s^2/2 i see so you assumed it was a right triangle. what if it is not?? coz it doesnt say..
it doesn`t matter. its like picking numbers. pick a triangle that fits for the statement to be true.
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Joined: 22 Aug 2005
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I guess question is asking to find raio of sides for two similar triangles.
if A and B are similar:
Area(a) / Area(B) = sidea^1/side2^2
putting value:
2/1 = s1^2/s2^2 or s1 = sqrt(s2) * 2



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Joined: 28 May 2005
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from the question my understanding is that one of the side of one triangle is s and another side of simillar triangle is S
if the triangles are simillar.
then area of triangle 1/ area of triangle 2 = sidea^1/side2^2
or 2/1 = s^2/S^2
so S = s/sqrt2
or you can re write is as (sqrt2/2) * s
answer is A
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Re: Geometry  Triangles [#permalink]
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10 Nov 2005, 11:52
Guys,
Sorry for the confusion, the actual question has two diagrams. A small triangle and a large triangle. Both triangles are similiar triangles. the smaller triangle has side 's' and the larger triangle has side 'S'. The are of the smaller triangle can be x, and the area of the larger triangle can be 2x.
I hope that clears any confusion. btw, the OA (as mentioned is C). I got C by picking number, is there some concept I'm missing that would make the question easier?
Thanks!
desiguy wrote: Given two similiar triangles, if the area of one triangle, side s, is twice the area of the second triangleside S, then in terms of s, S =
(sqrt2/2)s (sqrt3/2)s sqrt2s sqrt3s 2s
Pleae show your calcs
A



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Re: Geometry  Triangles [#permalink]
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10 Nov 2005, 12:22
desiguy wrote: Guys, Sorry for the confusion, the actual question has two diagrams. A small triangle and a large triangle. Both triangles are similiar triangles. the smaller triangle has side 's' and the larger triangle has side 'S'. The are of the smaller triangle can be x, and the area of the larger triangle can be 2x. I hope that clears any confusion. btw, the OA (as mentioned is C). I got C by picking number, is there some concept I'm missing that would make the question easier? Thanks! desiguy wrote: Given two similiar triangles, if the area of one triangle, side s, is twice the area of the second triangleside S, then in terms of s, S =
(sqrt2/2)s (sqrt3/2)s sqrt2s sqrt3s 2s
Pleae show your calcs
A
for similar triangles=> ratio of sides are equal => s/S = h/H
I am just naming the height of each triangle as h and H. since they are similar the heights will be in the same ratio.
2 x Area of small triangle = Area of big triangle
or, 2 x (1/2) s x h = (1/2) S x H
or, s x h = (1/2) S x H
or, S = 2s x h/H
or, S = 2s x s/S
or S^2 = 2 s^2
or S = sqrt(2). s




Re: Geometry  Triangles
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10 Nov 2005, 12:22






