# The Wire - Areas of a Square & Circle Along a Wire CONT'D

• Sep 26th 2012, 12:48 PM
Seshiru
The Wire - Areas of a Square & Circle Along a Wire CONT'D
Okay, so I've gotten a bit farther in this problem http://mathhelpforum.com/pre-calculu...long-wire.html

I know that the area of the square is ((360-x)/(4))^2 and I think the area for the circle is x^2/4pi. I'm not too sure about this, but I'll outline my process below.

SQUARE: Let 360-x represent the piece of the string formed into the square. In order to find the length of one side of the square, we need to divide this by 4. Then, we need to square it to get the area, so it turns out to be ((360-x)/(4))^2.

CIRCLE: This part was trickier for me. We know that 2(pi)(r) is the formula for the circumfrence, and (pi)(r)^2 is the formula for the area. So, we let x = the piece of string formed into the circle. We set x = 2(pi)(r) and get r = x/2pi(r). Then, we plug that back into the area formula to get x^2/4pi at the end of it all. Keep in mind, I'm not too sure about this one, so I may be wrong.

I'm not sure how to proceed. I know that I either set the two areas equal to each other, or I add them and set the whole thing equal to 360. Can anybody please direct me on how to proceed? I tried both methods and I get some whacky answer, so that leads me to believe that either my circle's area is wrong, or my algebra isn't correct. If someone could outline the steps for me, I would greatly appreciate it. Thank you!
• Sep 26th 2012, 01:43 PM
ebaines
Re: The Wire - Areas of a Square & Circle Along a Wire CONT'D
Quote:

Originally Posted by Seshiru
I know that the area of the square is ((360-x)/(4))^2 and I think the area for the circle is x^2/4pi. I'm not too sure about this, but I'll outline my process below.

This is correct.

Quote:

Originally Posted by Seshiru
I'm not sure how to proceed. I know that I either set the two areas equal to each other, or I add them and set the whole thing equal to 360.

The objective is to have the two areas equal, so set them equal to each other:

$\frac {(360-x)^2}{16} = \frac { x^2}{4 \pi}$

You'll have to rearrange this to get it into the quadratic form ax^2 + bx+ c = 0, and then apply the quadratic formula to solve for x:

$x = \frac {-b \pm \sqrt{ b^2-4ac}}{2a}$
• Sep 26th 2012, 02:12 PM
Seshiru
Re: The Wire - Areas of a Square & Circle Along a Wire CONT'D
Nevermind, I got it. Thanks a bunch man!