1. ## Frustrating Geometry Problem

Ok... I have to do this problem attached.

It's so frustrating. 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. 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. We know that x + y = 1. I then tried to find the hypotenuse of square 1, 2 (and consequently 3) as well as another one of 1.

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

2. 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
I assume that:

1. I have to show that square 3 is half the area of the entire square.

2. We know that x + y = 1

If so you know:

$\displaystyle y^2 = \frac12~\implies~y=\frac12 \cdot \sqrt{2}$ .........Length of side of square 3

$\displaystyle (1-x)^2=\frac12~\implies~x=1-\frac12 \cdot \sqrt{2}$.........Length of side of square 1

Consequently the side length of square 2 is:

$\displaystyle \frac12 \cdot \sqrt{2} - \left(1-\frac12 \cdot \sqrt{2}\right) = \sqrt{2} - 1$