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molt_llest: Senior | Next Rank: 100 Posts; Posts: 69; Joined: Sat May 10, 2008 3:05 pm; Thanked: 1 times

by molt_llest » Sun Jun 01, 2008 11:15 am

I thought this was pretty easy but I'm doing something wrong.

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Source: Beat The GMAT — Problem Solving | Sun Jun 01, 2008 11:15 am

kishore: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Fri Dec 21, 2007 2:58 pm; Location: Fremont , California; Thanked: 3 times

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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jasonc: Master | Next Rank: 500 Posts; Posts: 127; Joined: Sat Apr 19, 2008 1:02 am; Thanked: 12 times

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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chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

by chidcguy » Sun Jun 01, 2008 12:21 pm

How can you use the side proportionality to replace the height?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

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.

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

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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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

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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chidcguy: Master | Next Rank: 500 Posts; Posts: 438; Joined: Mon Feb 12, 2007 9:44 am; Thanked: 26 times

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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aj5105: Legendary Member; Posts: 1169; Joined: Sun Jul 06, 2008 2:34 amFollowed by: 1 member; Thanked: 25 times

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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canuckclint: Master | Next Rank: 500 Posts; Posts: 154; Joined: Tue Aug 26, 2008 12:59 pm; Location: Canada; Thanked: 4 times

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

Clinton
Math Minor

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