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# A thin piece of wire 40 meters long is cut into two pieces.

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Joined: 11 Jan 2006
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A thin piece of wire 40 meters long is cut into two pieces. [#permalink]  10 Aug 2006, 03:00
8. A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square regions in term of r?
A. (pie) r2
B. (pie)r2 +10
C. (pie)r2 + Â¼ (pie) 2 r2
D. (pie)r2+(40-2(pie)r) 2
E. (pie)r2 +(10-1/2 (pie)r) 2
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Joined: 07 Nov 2005
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Got E.
Let X be the first piece.
Second piece-----> 40-X.

Now, X is used to form a circle
=> X = 2*(pie)*r ,.....................Eqn(1) ,(Circumference = length of wire)

Area of Circle---------> (pie)*r^2.

Also,let side of square be a,
40 - X = 4*a , ........................(Perimeter (4*a) = Length (40 -X) )
Substituting X from Eqn(1) ,
=> 40 - {2*(pie)*r} = 4*a ,
=> 10 - { ((pie)*r) / 2 } = a.
Area of Square = a^2

Hence total area ==> (pie)r^2 +(10-1/2 (pie)r)^ 2
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Let one peice be x, hence the other peice is 40-x

x = 2pir

Hence 40-x = 40-2pir

Let second peice form the square
Hence 4S = 40 - 2pir
S = 10 - pir/2
Area of square = (10-pir/2)^2

Sum of areas = pir^2 + (10-pir/2)^2

Hence E
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