+0  
 
0
565
1
avatar

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?

Guest Apr 27, 2015

Best Answer 

 #1
avatar+17745 
+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 ]

geno3141  Apr 27, 2015
 #1
avatar+17745 
+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

15 Online Users

avatar

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.