Question

A piece of wire x inches long will be bent into the shapes below. Write an expression for the enclosed area of each shape in terms of x.

a) Circle: A=_______ square inches.
b) Square: A=_______ square inches.

Answer 1

Here you go. http://www.mathsisfun.com/area.html

Answer 2

@Jhannybean I'm still struggling with this.

Answer 3

I tried setting x equal to the formula of the circumference and then solving for r and substituting into the area formula.

Answer 4

One minute D:

Answer 5

@iamsammybear...what do you mean 'you tried'? Can you show your work please?

Answer 6

\[x=2 \pi r^{2}\]\[r=\sqrt{x-2\pi}\]\[A=\pi (x-2\pi)\]

Answer 7

you made a mistake while solving for r.

Answer 8

Your formula for circumference is also incorrect

Answer 9

\(x = C = 2\pi r\)
\(x = 2\pi r\)
\[r = \frac{x}{2\pi}\]

Answer 10

Alright. in order to find the Area of the circle in terms of x, first we need to fidn the circumference....since we're given a wire and we want to see HOW MUCH of the wire it would take to circumscribe the circle.

C = \(2 \pi r\)

Putting this in terns of x, we can say

\(x= 2 \pi r\) therefore

\(r= x/2\pi\)

Now that we've found our radius, we can plug it into the equation of a circle for Area.
\[\large A = \pi r^2\]\[\large A = \pi \left( \frac{x}{2\pi}\right)\]\[\large A = \frac{x}{2}\]

Answer 11

Oh, I see.
Thank you so much @Hero!

Answer 12

oh no! i forgot the squared.

Answer 13

@Jhannybean, you made a mistake in your simplifcation

Answer 14

Yes i know.....
\[\large A = \pi \left(\frac{x}{2\pi}\right)^2\]\[\large A = \pi \left(\frac{x^2}{4\pi^2}\right)\]\[\large A = \frac{x^2}{4\pi}\]

Answer 15

Thank you all so much!
I really appreciate the help from all of you.

Answer 16

And then for the square.. \[\large A = x = 4s \]\[\large s = \frac{x}{4}\] Now plug it into the formula for Area of a square, A = s^2 \[\large A = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}\]

Answer 17

@Jhannybean, (as a suggestion) you should have allowed @iamsammybear to try the other one on her own.

Answer 18

You are right....