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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

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 #1
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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
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1+0 Answers

 #1
avatar+17655 
+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

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