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If you have two similar triangles, one with an area twice as

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If you have two similar triangles, one with an area twice as [#permalink]

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New post 27 May 2006, 14:24
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If you have two similar triangles, one with an area twice as much as the other, what is the S in terms of s?

The smaller triangle has base "s". The larger triangle has base "S". They are not isosceles or equilateral.

(A) [s*sqrt(2)/2]
(B) [s*sqrt(3)/2]
(C) s*sqrt(2)
(D) s*sqrt(3)
(E) 2s

Please explain.

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Re: PS: Geometery (Similar Triangles) [#permalink]

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New post 27 May 2006, 14:29
x242222 wrote:
If you have two similar triangles, one with an area twice as much as the other, what is the S in terms of s?

The smaller triangle has base "s". The larger triangle has base "S". They are not isosceles or equilateral.

(A) [s*sqrt(2)/2]
(B) [s*sqrt(3)/2]
(C) s*sqrt(2)
(D) s*sqrt(3)
(E) 2s

Please explain.


C

the area is directly proportional to square of sides , if they are similar.

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New post 28 May 2006, 00:03
Let H be the height of larger triangle and h be height of smaller one.

Area of larger = 2*area of smaller
1/2*S*H = 2*(1/2*s*h)
=> S*H = 2*s*h ---(A)
For similar traingles ratio of sides/height is equal.

Let S/s = H/h = k

S = s*k & H = h*k
Substitute in A

s*k*h*k = 2*s*h
=> k^2 = 2
k = sqrt(2)

S/s = sqrt(2)
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New post 28 May 2006, 01:43
Thanks giddi77. Nicely explained.

Some proportionality rules were hidden.

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New post 28 May 2006, 06:10
giddi77 wrote:
Let H be the height of larger triangle and h be height of smaller one.

Area of larger = 2*area of smaller
1/2*S*H = 2*(1/2*s*h)
=> S*H = 2*s*h ---(A)
For similar traingles ratio of sides/height is equal.

Let S/s = H/h = k

S = s*k & H = h*k
Substitute in A

s*k*h*k = 2*s*h
=> k^2 = 2
k = sqrt(2)
S/s = sqrt(2)

nice work...

i also got C in pretty much similar way.

1/2*S*H = 2*(1/2*s*h)
S*H = 2*s*h
S/s = 2h/H
S/s = 2 s/S [because S/s = H/h]
S^2/s^2 = 2
S/s = sqrt(2)

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  [#permalink] 28 May 2006, 06:10
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