+0

# 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?

0
642
1

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?

Apr 27, 2015

### Best Answer

#1
+17747
+5

The area of a square with side length x is x2.

The area of a circle with radius (x - 1)  is Π(x - 1) = 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 ]

Apr 27, 2015

### 1+0 Answers

#1
+17747
+5
Best Answer

The area of a square with side length x is x2.

The area of a circle with radius (x - 1)  is Π(x - 1) = 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 ]

geno3141 Apr 27, 2015

### New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.