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

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

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10 Aug 2006, 03:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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10 Aug 2006, 03:21
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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10 Aug 2006, 03:30
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
10 Aug 2006, 03:30
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