A wire is 50 inches long. Where should the wire be cut so that it forms 2 pieces one made into a circle and one made into a square such that the area of the circle plus the area of the square in exactly 80 square inches?
+33615 "A wire is 50 inches long. Where should the wire be cut so that it forms 2 pieces one made into a circle and one made into a square such that the area of the circle plus the area of the square in exactly 80 square inches?"
Hmm!
I think your total area should be 800 square inches NOT 80 square inches.
Let x = side length of the square.
50 - x =Circumference of the circle.
[50 - x] / 2pi =Radius of the circle
x^2 + [(50 -x)/(2pi)]^2*pi = 800, solve for x
x =27.567 inches - one side of the square.
50 - 27.567 =22.433 inches - Circumference of the circle.
So that:22.433/2pi =3.57 inches - radius of the circle.
27.567^2 + [3.57^2*pi] =~800 sq.inches.
So, the 50-inch wire should be cut into 2 pieces:
27.567 inches - being one side of the square.
22.433 inches - being the circumference of the circle.
+33615 1. If x is the side length of the square then the circumference of the circle is 50 - 4x rather than 50 - x.
2. However, note that the maximum possible area that can be achieved by a piece of wire 50 inches long is when the whole wire is used to make a circle. This results in an area of \(\pi (\frac{50}{2\pi})^2 = 198.94... in^2\) i.e. much less than 800 in2.
