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# there are two triangles whose three vertices have same

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Joined: 30 Oct 2007
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there are two triangles whose three vertices have same [#permalink]

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02 Nov 2007, 17:49
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

there are two triangles whose three vertices have same angles (x, y, z). If area of one triangle is twice the area of the other triangle, and the base of the smaller triangle is s, and base of the larger triangle is S, then what is S, in terms of s?

√2/2 s
√3/2 s
√2 s
√3 s
2 s

Last edited by iwo on 02 Nov 2007, 19:08, edited 1 time in total.

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Director
Joined: 11 Jun 2007
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Re: triangles question- seems hard [#permalink]

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02 Nov 2007, 18:12
iwo wrote:
there are two triangles whose vertices have same angles. If area of one triangle is twice the area of the other triangle, and the base of the smaller triangle is s, and base of the larger triangle is S, then what is S, in terms of s?

√2/2 s
√3/2 s
√2 s
√3 s
2 s

vertices have same angles <-- that tells us we are dealing with equilateral triangles with 60 degrees for each angle, so the height will be split with the 30-60-90 degree and will be 1/2*base* √3

i get A (although not 100% sure) please confirm

area = 1/2bh, stem tells us base is big S
area = 1/2 S * (√3)/2 S

smaller triangle
area = 1/2bh, base is little s
area = 1/2 s * (√3)/2 s

since area for larger triangle is twice that of the smaller
than S * (√3)/2S = 1/2 s * (√3)/2s

solving for S (multiply inverse 2/√3 on both sides to clear LHS)

S^2 = √3/4 s^2 * 2/√3
S^2 = s^2 * 2/4
and take the sqrt on both sides:
S = s * √(1/2)
S = s * √2/2

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02 Nov 2007, 19:06
nope thats not the OA

both triangles have same angles x-y-z, no degrees are given, sorry i forgot to mention that

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02 Nov 2007, 19:21
iwo wrote:
nope thats not the OA

both triangles have same angles x-y-z, no degrees are given, sorry i forgot to mention that

duh!!! sorry i know where i screwed up (hope not on test day)

continuation...

since area for larger triangle is twice that of the smaller
than S * √3/4S = 2* (s * √3/4s)

solving for S (multiply inverse 4/√3 on both sides to clear LHS)

S^2 = √3/2 s^2 * 4/√3
S^2 = s^2 * 4/2
and take the sqrt on both sides:
S = s * √2
so the third choice.. . is that correct?

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Re: triangles question- seems hard [#permalink]

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02 Nov 2007, 23:53
iwo wrote:
there are two triangles whose three vertices have same angles (x, y, z). If area of one triangle is twice the area of the other triangle, and the base of the smaller triangle is s, and base of the larger triangle is S, then what is S, in terms of s?

√2/2 s
√3/2 s
√2 s
√3 s
2 s

Getting C.

The ratio of the area of similar triangles is the square of the ratio of similar sides.

Since all three angles are equal the 2 triangles are similar.

Ratio of ar. of small triangle : ratio of ar. of big triangle = 1:2
Therefore, ratio of sides of small triangle : ratio of sides of big triangle = 1 : sqrt(2)

If s = 1 then S = sqrt(2)s

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Re: triangles question- seems hard   [#permalink] 02 Nov 2007, 23:53
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# there are two triangles whose three vertices have same

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