View Full Version : some help please..

Danielle058

08-20-2005, 12:57 AM

Alright, this is a word problem that I'm having trouble figuring out; if anyone can help please do :D

A piece of wire 5 inches long is to be cut into two pieces. One piece is x inches long and is to be bent into the shape of a sqaure. The other piece is to be bent in the shape of a circle. Find and expression for the total area made up by the square and the circle as a function of x.

THANKS SO MUCH.

The equations you need are:

c = 2*pi*r circle circumfrence

a=pi*r² circle area

a=s² square area

solve the first for r

substitute r in the second

s = x/4 in the third

c = 5-x in the modified second

Denis

08-20-2005, 09:10 AM

Let pi = p ; if you follow Gene's instructions, you'll end up with:

Combined Area = [x^2(p + 4) - 20(2x - 5)] / (16p)

So keep "manipulating" Gene's stuff until you get that...

Danielle058

08-23-2005, 11:39 AM

How comes all those other equations are needed? I get lost when you talk about those other equations why can't you just do this. The original piece is 5 inces. If the first piece is cut into a square, and that piece is x inches, than the first area would simply be x^2 wouldn't it. The second peice would than be 5-x and the area for a circle being.. a=pi(r^2). so in if the length is put in as R woudn't be (X^2-10x+25)pi and than couldn't you add these together to come up with the total area which would be x^2+(x^2-10x+25)(pi). I dunno if this is right or not so if someone could please check this and see if what I am doing is right. If it's not could smeone explain to me how and why I use the equations that have already been mentioned. Thanks :P

wjm11

08-23-2005, 01:37 PM

The original piece is 5 inces. If the first piece is cut into a square, and that piece is x inches, than the first area would simply be x^2 wouldn't it. The second peice would than be 5-x and the area for a circle being.. a=pi(r^2). so in if the length is put in as R woudn't be (X^2-10x+25)pi

Danielle,

You got it partly right; the square is x^2 in area, and the remaining piece is 5-x long. however, the 5-x is NOT the radius of the circle; it is the circumference of the circle. You need to use

5-x = 2*pi*r

Rearrange this to solve for r, then plug it into

A = pi*r^2

hope that helps.

But the square is (x/4)² in area. You divide the x by 4 to get the side of the square.

You are using my equations whether you know it or not.

------------------

Gene

wjm11

08-23-2005, 04:16 PM

But the square is (x/4)² in area. You divide the x by 4 to get the side of the square.

Thanks, Gene. I wasn't paying close enough attention.

Powered by vBulletin® Version 4.2.2 Copyright © 2014 vBulletin Solutions, Inc. All rights reserved.