Triangle - GMATPrep CAT

[myfish]

I keep trying this one. Can anyone please give me a hint? Thanks

[Anurag@Gurome]

Refer to the post here >> https://www.beatthegmat.com/area-members ... tml#377833

[eagleeye]

myfish:

These are two similar triangles (since angles of both triangles are the same).
Now the thing with similar triangles that you need to remember is:
Ratio of all corresponding dimensions is the same.

In other words, if there are two similar triangles with vertices abc, and the other being ABC, then in that case:
ab/AB=bc/BC=ac/AC=h/H (where h and H are their corresponding heights).

Now in the problem at hand:
s/S = h/H (where i have assumed height of smaller triangle at angle y being h, and height of larger triangle at angle y being H).
Then
Area of smaller one = 1/2*s*h
Area of larger one = 1/2*S*H
Ratio of smaller to larger = (s*h)/(S*H), But we know that h/H=s/S, so Area(smaller)/Area(larger)=s^2/S^2

We are given that Area(smaller)/Area(larger) = 1/2
therefore 1/2 = s^2/S^2, so S=s*sqrt(2).

This was the complete math solution. Now here's the faster way of doing it, so that we can save time on the GMAT.

You need to remember that for similar triangles:
1. Ratio of corresponding dimensions is equal.
2. Ratio of areas is square of the ratio of corresponding dimensions.

Then we will have
Area of larger/Area of smaller = 2 = S^2/s^2 (using point 2)
Then S= s*sqrt(2)

Let me know if this helps

[dhonu121]

Since, the angles are not mentioned and the only fact that we know is that triangles are similar. So, for covenience, assume that both the triangles are equilateral. One triangle is just a scaled version of other.
Hence area(s)=area(S) or 2*sqrt(3)/4*s^2 = sqrt(3)/4*S^2
or S=sqrt(2)*s.
Hence C