If an equilateral triangle and a square have the same area, what is the ratio of their perimeters?
by the way, my math teacher is having us email a bum and he sends us problems, if we get it right 1% of extra credit added to our grade. I have 99% in that class so... whatever :), I'll pass the problems onto you guys as they come. This is problem #1!
seriously a bum?
Yeah, pretty much. He doesn't have a job and lives in an RV, but he has money from relatives to spend etc. To get internet he jacks it from other people's wireless connection.
1:(2/3)*3^(1/4)
Well, if x is the length of one side of the triangle, it's:
3x:2x*3^(1/4)
That's a wild guess from what I remember from geometry.
I get this: (http://latex.sidoh.org/?render64=XGRpc3BsYXlzdHlsZSBcZnJhY3tcc3FydHs1fSs0fXs2fQ==), but I have a creeping suspicion that it's supposed to be the golden ratio...
by plugging in numbers I got 1.139753528:1 :P, but I can't replicate that algebraically
Ok, here goes how I did it.
If x is the length of the side of the triangle, the height is sqrt(3)*x/2. So, the area is now: 1/2*sqrt(3)*x/2*x=(sqrt(3)*x^2)/4.
All right, so now that the areas have to be equal, let y be the length of the side of a square. y^2=(sqrt(3)*x^2)/4, so the length of the side is: y=(3^(1/4)*x)/2
Ok, so now we have the perimeters. 3x for the triangle, and 4y or 4*(3^(1/4)*x)/2 or 2x*3^(1/4) for the square.
3x:2x*3^(1/4).
edit:
Quote from: Towelie on March 31, 2007, 12:35:35 PM
by plugging in numbers I got 1.139753528:1 :P, but I can't replicate that algebraically
If you put a number into mine and simplify, you will get 1:0.877383, and 1/0.877383=what you got.
Yep, that's right. Working out the problem after reading it correctly (and realizing I found the height of the triangle incorrectly... lol), I got the same thing.
I probably should have eaten something before I tried that, even as trivial is it is... lol