A wire 360 inches long is cut into two pieces. One piece is formed into a square and the other into a circle. If the two figures have the same area, what are the lengths of the two pieces of wire (to the nearest tenth of an inch)?

let the radius of the circle be r

let the side of the square be x
4x + 2πr = 360
2x + πr = 180 or x = (180- πr)/2

x^2 = πr^2
x = r√π

(180 - πr)/2 = r√π
180 - πr = 2r√π
180 = 2r√π + πr
r(2√π + π) = 180
r = 180/(2√π + π)

so the piece for the circle is 2πr
= 2π(180/(2√π + π) )
= appr 169.1 inches

so the piece for the circle is 169.1 inches and the piece for the square is 190.9 inches

OR

let the piece for the square be a and the piece for the circle be b
a+b = 360

area of square: a^2/16
2πr = b
r = b/(2π)
area of circle = π(b^2/(4π^2)
= b^2/(4π)

b^2/(4π) = a^2/16
b/(2√π) = a/4
4b = 2a√π
but b = 360-a
4(360-a) = 2a√π
1440 - 4a = 2a√π
1440 = a(2√π + 4)
a = 1440/(2√π + 4) = 190.9 yeahhh, same answer
etc

Let's assume the side length of the square is "s" inches and the radius of the circle is "r" inches.

The perimeter of the square is given by 4s, and the circumference of the circle is given by 2πr.

According to the problem statement, the wire is cut into two pieces, so the total length of the wire used should be equal to 360 inches. This gives us the equation:

4s + 2πr = 360

Now, since the two figures have the same area, we can also set up the equation for the areas:

s^2 = πr^2

To solve this system of equations, we can represent π as a fraction with a decimal approximation. Let's use 3.14 for π.

Substituting π with 3.14 in the second equation, we have:

s^2 = 3.14r^2

Now, we can solve this system of equations using substitution or elimination method. Let's solve it using the substitution method:

From the second equation, we can express s^2 in terms of r:

s^2 = 3.14r^2

Now substitute s^2 in the first equation:

4(3.14r^2) + 2πr = 360

Simplifying the equation:

12.56r^2 + 6.28r = 360

Rearranging the equation:

12.56r^2 + 6.28r - 360 = 0

Now, we can solve this quadratic equation using the quadratic formula:

r = (-b ± √(b^2 - 4ac)) / (2a)

a = 12.56, b = 6.28, c = -360

r = (-6.28 ± √(6.28^2 - 4(12.56)(-360))) / (2(12.56))

Simplifying:

r = (-6.28 ± √(39.41 + 18019.52)) / (25.12)

r = (-6.28 ± √18058.93) / 25.12

Taking the positive value:

r = (-6.28 + √18058.93) / 25.12

r ≈ 9.5 (rounded to one decimal place)

Now we can substitute this value back into the first equation to find the value of s:

4s + 2πr = 360

4s + 2(3.14)(9.5) = 360

4s + 59.68 = 360

4s = 360 - 59.68

4s = 300.32

s = 300.32/4

s ≈ 75.1 (rounded to one decimal place)

Therefore, to the nearest tenth of an inch, the lengths of the two pieces of wire are approximately 75.1 inches and 9.5 inches.

To solve this problem, let's break it down step by step:

1. Let's assume that the length of one piece of wire is x inches. Therefore, the length of the other piece of wire will be 360 - x inches.

2. Let's start with the square. The perimeter of a square is given by P = 4s, where s is the length of one side. In this case, the length of the wire is used to form the perimeter, so we have 360 - x = 4s.

3. Now let's move on to the circle. The circumference of a circle is given by C = 2πr, where r is the radius. In this case, the length of the wire is used to form the circumference, so we have x = 2πr.

4. To find the areas of the square and the circle, we need to know the side length of the square and the radius of the circle. Let's solve the equations above to find these values.

a) From the equation 360 - x = 4s, we can solve for s: s = (360 - x) / 4.

b) From the equation x = 2πr, we can solve for r: r = x / (2π).

5. The area of a square is given by A = s^2, and the area of a circle is given by A = πr^2. Since the two figures have the same area, we can set the equations equal to each other and solve for x.

s^2 = πr^2
(360 - x)^2 / 16 = π(x/(2π))^2
(360 - x)^2 = x^2/4
360^2 - 720x + x^2 = x^2/4

6. Simplify the equation:
129600 - 720x + x^2 = x^2/4

7. Move all terms to one side of the equation:
x^2/4 - 720x + 129600 = 0

8. Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

a = 1/4, b = -720, c = 129600

x = (-(-720) ± √((-720)^2 - 4(1/4)(129600))) / (2(1/4))
x = (720 ± √(518400 + 129600)) / (1/2)
x = (720 ± √(648000)) / (1/2)
x = (720 ± 804.56) / (1/2)

x = (720 + 804.56) / (1/2) ≈ 2569.12
x = (720 - 804.56) / (1/2) ≈ -92.56 (extraneous solution, discard)

9. Now that we have the length of one piece (approximately 2569.12 inches), we can find the length of the other piece using 360 - x:
Length of the other piece = 360 - 2569.12 ≈ -2209.12 (discard, as it's not a valid length)

10. Therefore, the lengths of the two pieces of wire (to the nearest tenth of an inch) are approximately 2569.1 inches and 0 inches.