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GMAT Prep Geometry Question.

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by pepeprepa » Tue Jul 22, 2008 8:11 am
Well First triangle (area A) can be included in the Second triangle (area B).

Each of the ratio between the considering sides are equal.
So s/S is a ratio that can make you know that s/S=t/T=k/K if we call the sides of the first triangle s,t,k and the sides of the second one S,T,K
This ratio is the same for the heights of the triangles, let's call them h and H which make 2 new triangles with same angles (it would be better if I could show you with a drawing, it's long to explain but it takes you 30" if you know the theorems)
So we have s/S=h/H

Now with information of the question
2A=B
2*((s*h)/2)=(S*H)/2
2*s*h=S*H
2*s*(h/H)=S
2*s*(s/S)=S
2*(s^2)=S^2
S=s*sqrt(2)

It's ok.
Just draw if you do not catch and read the theorem I used.
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by sudhir3127 » Tue Jul 22, 2008 8:52 am
u need to use the following theorem...

Side-Side-Side (SSS) theorem: If two triangles have all three pairs of corresponding sides in the same ratio, then the triangles are similar. In other words, if triangles ABC and A'B'C' have A'B' / AB = B'C' / BC = C'A' / CA , then the triangles are similar

thats how u get h/h = s/S
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by anju » Tue Jul 22, 2008 2:51 pm
The ratio of the areas of 2 similar triangle is the square of the ratio of corresponding lengths:

Area (bigger triangle)/ area(smaller triangle) = (S/s)^2
As per the information provided, cosider area (smaller triangle) = x then
2x/x = (S/s)^2
2 * s^2 = S^2
sqrt2 * s = S
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by gmatutor » Tue Jul 22, 2008 5:03 pm
Since you are not given any angles you, this must be true for all triangles. You are not told that angle y=x, you are not told that they are not equal.

Look at this as two right isosceles triangles. Then you can see that the hypotenuse of the smaller triangle is equal to side of the larger triangle.

The ratio of a side of a triangle to the hypotenuse of a triangle is 1:sqrt(2), so S must equal s*sqrt(2).
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