GMAT Prep Trig Question
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Hi guys, please help me out with this one, I'm not sure how to approach it. Thanks.
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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.
area of square = s ^ 2.
Now sqrt(3) t^2 / 4 = s ^ 2
so t : s = ...
So IMO D.
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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we want to get t/s, so we rearrange to that point.its_me07 wrote:how do u solve this
t^2(sqrt3)/4=s^2 for t:s??????
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]
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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.
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.