Originally Posted by
Startbucks ... I have to find the length of squares 1, 2 and 3. Also, I have to show that square 3 is twice the area of the entire square. To me that's impossible: The included square can't be greater than the entire square. I'd like to share some of my work.
I know each side is 1, so we know the diagonal is sqrt(2).
I let the side of the square 1 be x. I let the side of square 3 be y-x. According to your sketch this is the side of square 2. We know that x + y = 1. Where from do you know this? I then tried to find the hypotenuse of square 1, 2 (and consequently 3) as well as another one of 1. Do you mean diagonal? Why?
So 2*(hypot. of square 1) + hypot. of square 2 = sqrt(2)
So, sqrt(2*(y-x)^2) + sqrt(2*sqrt(2x^2)) = sqrt(2)
But when I solve this I get such a complicated answer and I'm so stuck.
I also tried using proportions, like saying:
1/sqrt(2) = x/(sqrt(2x^2)) ... but all my calculator gave me was x > 0 which wasn't helpful.
Any help would be great