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27 May 2006, 15:24
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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.
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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28 May 2006, 01:03
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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)
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)
