GMATPrep2 - Geometry ps

Question

Please explain your answer. I used right triangle as my prototype, but that didn't work.

GMATQuantCoach3 · Answer

This is a similar triangle problem.

raito of the areas = (ratio of the sides)^2

2 = (ratio of the sides)^2

ratio of the sides = sqrt(2)

C is the answer.

coelholds · Answer

Seofah

Is the answer 2s, or is just your option? in my calculations I got  S = sqrt{2}*s , but I did it very fast, so I am not sure.

coelholds · Answer

Ok, someone has also posted at the same time.

So here is what I did, although the first solution is faster if you assimilate things easily.

I have used similarity:

H = Highest height
h = lowest height

S/s = H/h 
So  h/H = s/S

The area of the second is two the area of the first, so
 S*H/2 = s*h/2 
 S=2*s*h/H 
 S=2*s*s/S 
 S=sqrt{2}*s

superman · Answer

Let us take Area of right triangle as A1 and left as A2

A1 = 2A2
1/2*S*H = 2*(1/2*s*h )

1/2*S*H= s*h

S*H = 2*s*h .... (1)

Now note that angle are same for both the triangles therefor they are similar triangles

h/H = s/S

h=sH/S

Using the above equation in equation 1 we have

S*H=2*s(sH/S)

Solving it we get S=\sqrt{2} s

Answer is C

Can you please confirm ?

irajeevsingh · Answer

Nicely explained GMATQuantCoach3 +1 Kudo

seofah · Answer

OA C.
Thanks for the effort!

Prep Toolkit