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singhmaharaj: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Tue Feb 10, 2009 6:24 amFollowed by: 2 members; Location: Hong Kong

by singhmaharaj » Tue May 06, 2014 2:16 am

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue May 06, 2014 6:27 pm

Whenever the triangles are similar then,

(Area of Triangle A) : (Area of Triangle B) = (side of A)^2 : (side of B)^2

So,
A : 2A = s^2 : S^2

==> S^2 = 2 * s^2
==> S = sqrt(2) * s

[spoiler]{C}[/spoiler]

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singhmaharaj: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Tue Feb 10, 2009 6:24 amFollowed by: 2 members; Location: Hong Kong

by singhmaharaj » Wed May 07, 2014 5:10 am

Hi,

As far as I know, the property is

if two similar triangles have corresponding lengths in the ratio a:b, then their areas will be in the ratio (a)^2 : (b)^2

How did you derive the property other way round??

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Wed May 07, 2014 5:14 am

singhmaharaj wrote:Hi,

As far as I know, the property is

if two similar triangles have corresponding lengths in the ratio a:b, then their areas will be in the ratio (a)^2 : (b)^2

How did you derive the property other way round??

Are you about this property? I doubt..

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singhmaharaj: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Tue Feb 10, 2009 6:24 amFollowed by: 2 members; Location: Hong Kong

by singhmaharaj » Wed May 07, 2014 5:21 am

That is how I understood it from Manhattan GMAT Geometry section on similar triangles.

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Wed May 07, 2014 11:39 pm

singhmaharaj wrote:That is how I understood it from Manhattan GMAT Geometry section on similar triangles.

Did you find this information somewhere online? If so, then can you please share the link.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu May 08, 2014 1:59 am

When two triangles are similar, the length of EACH SIDE of the larger triangle is some factor (multiplier) greater than its corresponding side of the smaller triangle.
So, if we say that the length of each side of the larger triangle is k TIMES the length of each corresponding side of the triangle, we can write S = sk
Likewise, if we let H = the height of the large triangle and let h = the height of the small triangle, then we can write H = hk

We're told the following: (area of large triangle) = 2(area of small triangle)
Area = (1/2)(base)(height), so we can write: (1/2)SH = 2(1/2)sh
Divide both sides by 1/2 to get: SH = 2sh
Now substitute to get: (sk)(hk) = 2sh
Simplify: shkÂ² = 2sh
Divide both sides by sh to get: kÂ² = 2
Solve, to get k = âˆš2

Great! So, the length of each side of the larger triangle is âˆš2 the length of each corresponding side of the triangle.

So, the length of side S = (âˆš2)s

Answer: C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Mon May 12, 2014 11:47 am

singhmaharaj wrote:Hi,

As far as I know, the property is

if two similar triangles have corresponding lengths in the ratio a:b, then their areas will be in the ratio (a)^2 : (b)^2

This isn't quite right.

Suppose that a = 6 and b = 2, and further suppose that the height of triangle A is 12 and the height of triangle B is 4. The area of A is thus (6*12)/2, or 36, and the area of triangle B is thus (2*4)/2, or 4.

Notice that the larger triangle has an area equal to (a/b)Â² times the area of the smaller triangle. This will always be true, as we can demonstrate algebraically.

Suppose that B is the smaller triangle, and that B has a base of b units and a height of h units. Since A and B are similar, A's base = b*k, where k is some number greater than 1, and A's height = h*k, where k is the same multiplier that it was before (since the triangles are similar).

Then the area of A = (bk)(hk)/2, and the area of B = (b*h)/2. Thus A has an area equal to the area of B times kÂ², or the square of the ratio between the two corresponding bases.

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