Ratio question

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by DanaJ » Tue Mar 10, 2009 7:52 am
The area of the equilateral triangle will be [(t^2)sqrt(3)]/4.
The area of the square will be s^2.
The two are equal, so you get that:
[(t^2)sqrt(3)]/4 = s^2
(t^2)sqrt(3) = 4(s^2) - apply sqrt andd get:
t*(fourth degree root of 3) = 2s.

This makes t = 2s/(fourth degree root of 3) and t:s 2:(fourth degree root of 3).

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by zephir21 » Tue Mar 10, 2009 8:07 am
Hi

You have Area1 = Area2

t/2*(sqr(t^2-(t/2)^2))=s^2

t/2*sqrt(3/4*t^2)=s^2
sqrt(3)/4*t^2=s^2

Put sqrt to the whole equation

sqrt^4(3)/2*t=s

You have your answer ;)