A square has side length x and a circle has radius (x - 1). At what value of x will the two figures have the same area?
The area of a square with side length x is x2.
The area of a circle with radius (x - 1) is Π(x - 1)2 = Π(x - 1)(x - 1) = Π(x2 - 2x + 1) = Πx2 - 2Πx + Π
Since they are equal: x2 = Πx2 - 2Πx + Π
Getting all terms to one side: 0 = Πx2 - x2 - 2Πx + Π
Factoring: 0 = (Π - 1)x2 - 2Πx + Π
Using the quadratic formula with a = (Π - 1), b = -2Π, and c = Π
x = [2Π ± √( (-2Π)2 - 4(Π - 1)(Π) ) ] / [ 2(Π - 1) ]
x = [2Π ± √( 4Π2 - 4Π2 + 4Π ) ] / [ 2(Π - 1) ]
x = [2Π ± √(4Π) ] / [ 2(Π - 1) ]
x = [2Π ± 2√(Π) ] / [ 2(Π - 1) ]
x = [Π ± √(Π) ] / [ Π - 1 ]
So, either x = 0 or x = [Π + √(Π) ] / [ Π - 1 ]
The area of a square with side length x is x2.
The area of a circle with radius (x - 1) is Π(x - 1)2 = Π(x - 1)(x - 1) = Π(x2 - 2x + 1) = Πx2 - 2Πx + Π
Since they are equal: x2 = Πx2 - 2Πx + Π
Getting all terms to one side: 0 = Πx2 - x2 - 2Πx + Π
Factoring: 0 = (Π - 1)x2 - 2Πx + Π
Using the quadratic formula with a = (Π - 1), b = -2Π, and c = Π
x = [2Π ± √( (-2Π)2 - 4(Π - 1)(Π) ) ] / [ 2(Π - 1) ]
x = [2Π ± √( 4Π2 - 4Π2 + 4Π ) ] / [ 2(Π - 1) ]
x = [2Π ± √(4Π) ] / [ 2(Π - 1) ]
x = [2Π ± 2√(Π) ] / [ 2(Π - 1) ]
x = [Π ± √(Π) ] / [ Π - 1 ]
So, either x = 0 or x = [Π + √(Π) ] / [ Π - 1 ]