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by kishore » Sun Jun 01, 2008 12:03 pm
Answer: C

A(S) = 2 A(s)

= 2 *( 1/2 * s * h) ( area = 1/2 * base * height)

A(S) = 1/2 * S * H


From above, we can conclude that if S = sqrt(2) * s and H = sqrt(2) *h
then the A(S) = 2 A(s)

There fore S = sqrt(2) * s
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by jasonc » Sun Jun 01, 2008 12:11 pm
the area is twice as large does not mean the length of the sides are twice as long.

what is given is the following:
similar triangles = proportions of sides on the two triangles are equal
i.e. b/B=s/S=any side of the smaller triangle/the same side of the larger triangle

area of big = 2x area of small => (1/2*s*b)*2 = 1/2*S*B
=>s*b=1/2*S*B
=>2*s*b/B=S
=>2*s*(s/S)=S
=>2s^2=S^2
=>S = SQRT(2)*s
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by chidcguy » Sun Jun 01, 2008 12:21 pm
How can you use the side proportionality to replace the height?
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by Stuart@KaplanGMAT » Mon Jun 02, 2008 9:42 am
The key fact to remember to answer this question quickly is that area is a square function and length is a linear function.

We have similar triangles, which means that one is simply a bigger copy of the other; in other words, everything is proportional from one triangle to the next.

If we were told that the bigger triangle had LENGTHS twice the size of the smaller, then the area of the larger (since it's a square function) would simply be 2^2 = 4 times as big. If the bigger had lengths three times the size of the smaller, then the area of the larger would be 3^2 = 9 times as big.

However, we're not going from length to area, we're going from area to length... so instead of squaring, we need to square root.

So, if the area is twice as big, then each linear dimension will be sqrt2 as big: choose (sqrt2)s.
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by chidcguy » Mon Jun 02, 2008 10:07 am
Stuart,

Are you saying that the height of the larger triangle is also double the height of the smaller triangle, because its side is twice the side of the smaller triangle?
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by Stuart@KaplanGMAT » Mon Jun 02, 2008 7:12 pm
chidcguy wrote:Stuart,

Are you saying that the height of the larger triangle is also double the height of the smaller triangle, because its side is twice the side of the smaller triangle?
First, it's side is NOT twice the side of the smaller triangle.. it's side is sqrt2 * the side of the smaller triangle.

And yes, the height of the larger triangle is also sqrt2 * the height of the smaller triangle.
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by chidcguy » Mon Jun 02, 2008 7:41 pm
Thanks Stuart.

How can we say that the height of similar triangles are also proportional? AFAIK, all the material I have looked at (Kaplan,MGMAT) only pointed out that angles are equal and all the sides are in same proportion. If its mandatory that the heights of similar triangles are in same proportion as the sides are, why did every one miss this?

I looked for mathematical theorem or proof that says similar triangles have same height. I could not find any
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by aj5105 » Wed Oct 08, 2008 10:18 pm
IMO this is how it goes...

First,as look at the problem,we get to know the triangles are similar triangles.

For similar triangles -- square of the ratio of the sides = Ratio of the area of the triangles.

(s/S)^2 = As/AS

Now it's been given in the problem that AS=2As

Substitute-- we get s^2/S^2=1/2
Cross multiple-- 2s^2 = S^2
Therefore S = sqrt2 s
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by canuckclint » Sat Oct 11, 2008 10:33 pm
chidcguy wrote:Thanks Stuart.

How can we say that the height of similar triangles are also proportional? AFAIK, all the material I have looked at (Kaplan,MGMAT) only pointed out that angles are equal and all the sides are in same proportion. If its mandatory that the heights of similar triangles are in same proportion as the sides are, why did every one miss this?

I looked for mathematical theorem or proof that says similar triangles have same height. I could not find any
This one is easy to prove mathematically:

Take triangle 1 and 2 to have angle x and 90degree angle
Then we know for smaller triangle

tan(x) = h div s
For bigger triangle
tan(x) = H div S
Equate the two equations:
h div s = H div S
Height to side ratios are proportional.

z

x 90

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