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dauerbach
04-21-2009, 03:36 PM
OK, I don't even know where to start.

A 10-foot piece of wire is cut into two pieces. One piece is bent to make a square. The other forms a circle inscribed in the square. What is the length of the side of the square to the nearest tenth of a foot?

CookingWithGas
04-21-2009, 03:41 PM
You do know that the board prohibits doing your homework for you, right? I'm going to give you a hint instead of a solution, in case this is homework.

Set up an equation where the side of the square is x. The diameter of the circle is the same as the side of a square. So

10 = perimeter of the square + circumference of the circle

You should know how to express the perimeter of the square in terms of x, and the circumference of the circle in terms of x (its diameter).

Set up the equation and solve for x to the nearest tenth.

Thudlow Boink
04-21-2009, 03:44 PM
I hope this isn't a homework question.

One approach is to take a guess, then check to see if your guess is right. So, let's suppose the piece that's used to make the square is 6 feet and the other is 4 feet. (Draw a picture of a circle inside a square, and you should be able to see that the circle wire has to be shorter than the square wire.

How long is each side of the square?6/4, or 1.5 ft.

How long is the diameter of the circle?Since the circumference is 4 ft, and C = pi x d, the diameter is 4/pi ft, which is around 1.27

If the circle fits exactly inside the square, these would be the same length. Are they? No.

Now, here's where algebra comes in. Let x stand for the length of the wire used for the square. Then how long is the other piece, used for the circle? 10-x
Do the same thing with these quantities that you did for 6 and 4, set the side of the square equal to the diameter of the circle, and solve for x. (If you do it this way, your final answer isn't x, but x/4

dauerbach
04-21-2009, 03:58 PM
Ok, I get 10 = 4x + (x)(pi)
10/x = 4 + pi
10/x = 7.14
1/x = .714
x = 1/.714 =1.4 feet
That seems to be the correct answer.

Thudlow Boink
04-21-2009, 04:06 PM
That looks right dauerbach. Your way of setting up the equation is simple and elegant, probably more so than the way I was thinking about it. To solve it, I would have just combined the x terms (i.e. factored it as (4 + pi)x ) and divided both sides by (4 + pi)—which gives the same answer you got.

dauerbach
04-21-2009, 04:39 PM
I screwed it up.

The original length is 20 feet, not 10 feet.
I also assumed that one length of the square is a diameter, which it is not.

dauerbach
04-21-2009, 04:43 PM
No, no, no. The original length is 10 feet. But a length of the square is still not a diameter, right?

dauerbach
04-21-2009, 04:53 PM
So the original equation should be:

10 = 4x + sqrt (x^2) + (x^2) * (pi)
How do I solve that equation

Varrius
04-21-2009, 05:08 PM
No, no, no. The original length is 10 feet. But a length of the square is still not a diameter, right?

If the circle is inscribed inside the square so that the circle touches the square at four points that are exactly at the midpoints of each leg of the square, then the diameter of the circle would, in fact, equal the length of each leg of the square.

MikeS
04-21-2009, 10:34 PM
So the original equation should be:

10 = 4x + sqrt (x^2) + (x^2) * (pi)
How do I solve that equationYou were right the first time. The length of one side of the square is the diameter of the circle (divide the square into four equal smaller squares to see this more intuitively), and if the diameter of the circle is x then its circumference is pi*x.

CookingWithGas
04-23-2009, 07:00 AM
No, no, no. The original length is 10 feet. But a length of the square is still not a diameter, right?You were right the first time. If you draw a picture you will clearly see that the circle's diameter is the same as the side of the square.

Bosstone
04-23-2009, 08:47 AM
If the circle is inscribed inside the square so that the circle touches the square at four points that are exactly at the midpoints of each leg of the square,Interestingly (to me), this is exactly what inscribed (http://www.mathwords.com/i/inscribed_circle.htm) means in a math context. I hadn't known the definition was specific until I read this thread and went to look it up. I originally thought it just meant 'the circle is inside the square'.

Edit: Well, not exactly what it means. An inscribed circle doesn't have to touch the midpoints, like the linked diagram shows. But when dealing with a regular polygon like a square, it will.