GMAT Prep Trig Question

davidforsberg, IMO D. Because, area of the equ. trainagle = sqrt(3) t^2 / 4.
area of square = s ^ 2.
Now sqrt(3) t^2 / 4 = s ^ 2
so t : s = ...

So IMO D.

how do u solve this
t^2(sqrt3)/4=s^2 for t:s??????

its_me07 wrote:how do u solve this
t^2(sqrt3)/4=s^2 for t:s??????

we want to get t/s, so we rearrange to that point.

t^2(sqrt3) = 4(s^2)

(t^2)/(s^2) = 4/sqrt3

t/s = sqrt(4/sqrt3)

t/s = 2/(4throot3)

choose (d)

[note: we don't worry about the negative solutions because in geometry they make no sense - if this were an algebra problem we would have had multiple answers]

Hi Everyone,

I've solved this problem as follows:

1) area square = s^2
2) area triangle = 1/2*b*h
3) and s^2=1/2*b*h

Main issue (and time consuming part) for me was to work out/simplify the expression for the area of the triangle and some knowledge of exponents:

4) b=t
5) find h with pythagoras: t^2=h^2+{(1/2)*t}^2 --> h^2=(1-1/4)*t^2, thus, h=t*sqrt(3/4) = t*sqrt(3)*(1/2)
6) plug it back in 3) gives: s^2=(1/2)*t*t*sqrt(3)*(1/2)=sqrt(3)*(1/4)*t^2
7) simplify further and then take root: t/s=2/sqrt(sqrt(3), answer is D.

Hope this helps. Cheers, Joe.