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nunny
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gmatprep - geometry question 13 Jul 2008, 03:53
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Is there a way to solve this problem without using trigonometry?

pls refer to attachment.



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gmatprep - geometry question 13 Jul 2008, 04:33
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Let say height of the smaller triangle is h...and that of the larger triangle is H.
So s/S=h/H=k where K is a constant (as per rules of similarity)...so s=KS and h=kH
Now 1/2*s*h=1/2*1/2*S*H
ie...s*h=1/2*S*H
ie...KS*KH=1/2*S*H
ie...K^2 =1/2
ie...K=1/(sqrt 2)
ie...s/S = 1/(sqrt 2)
so S=(Sqrt 2)s
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gmatprep - geometry question 13 Jul 2008, 11:18
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1/2*S*H = 2*(1/2*s*h)
S = 2*s*h/H--------------A [Note: H & h are the respective heights for triangles with sides S and s]

We have similar triangles, so: h/s = H/S => h/H = s/S---B

Solve (A) and (B):
S = 2 * s * s/S
S^2 = 2 s^2
S = sqrt(2) s
C?
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gmatprep - geometry question 13 Jul 2008, 14:06
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While the two previous posts provide the correct answer, I'm going to provide a bit more narration through my answer to hopefully help others. I saw this on the GMATPrep 2 and got it wrong but it's a much simpler problem than it appears to be.

We're asked to find in terms of \(S\). We need to remember this for later.

If we make the height of s the variable \(a\), and let the height of S be \(b\) (similar to making it h and H)

Lets use a formula for the area of each triangle.

s = 1/2 * s * a

S = 1/2 * S * b

Now we also know that the proportion of a:b is the same as s:S. (This will be important later)

so to make an equation where we can solve for \(S\), we need get S on one side. To do this, we must make it 2s = S because s is 1/2 the area of S. It's not as simple to just answer E on the problem 2s because we've not taken into account the values of the height of each triangle.

Now substitute in from above

2 * 1/2 * s * a = 1/2 * S * b
2sa = Sb (the 1/2 on each side cancels out)

now, if we divide each side by Sa we get

2s/S = b/a

We know that a:b or a/b = s/S, so b/a = S/s, we can substitute these values in.

2s/S = S/s, now get a single s on one side and you have your answer. To get rid of the fractions, we can cross multiply.

\(2s^2 = S^2\)

Take the square root of both sides
\(\sqrt{2}s = S\)



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