Question

A wire 45 inches long is cut into two pieces. One piece is bent into the shape of a circle an the other a square. If the two figures have the same area, what are the lengths of the two pieces of wire?

Answer 1

A=pi r^2
A= side times side

tough one...

Answer 2

Let the legth of the first piece = x
So, the length of second piece is 45 - x

Answer 3

what do i do after that?

Answer 4

You will then express the area of the circle, say the first piece, in terms of x, using the formula for the area of the circle.

The second piece, which has length 45 - x, has area (45 - x)^2, becuase the area of a square with side s, is s^2. So the area of the square with side length 45 - x is (45-x)^2.

Answer 5

would i then set them equal to eachother?

Answer 6

Yes, the problem states that both have the same area.

Answer 7

or what will the equaiton look like so i can solve it?

Answer 8

You write the equation. You know one side of the equation is (45 - x)^2, the area of the square.

Answer 9

Now, you must write the area of the circle piece of wire.

Answer 10

so (45-x)^2 = pir^2

Answer 11

Hint: The circumference of the circle, C, is 2pi(r). And the circumference of the the wire is x.

Answer 12

so you have x = 2pi(r); so r, the radius = x/2pi

Now use that value of x, 2pi/r, and plug that into the formula for the area of the circle, pi(r^2)

Answer 13

Hence, your area of the circle is pi(x/2pi)^2

Answer 14

Now, you have both sides of your equation. Go solve for x. Good luck!

Answer 15

\[(45-x)^{2} = \pi (x/4\pi)^2\]

Answer 16

is that right?

Answer 17

Yes

Answer 18

Now solve for x. It isnt that difficult. First simplify the right side of the equation.

Answer 19

(45−x)2=(x/4)^2