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I got it....But some where i did some mistake...Pls help

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I got it....But some where i did some mistake...Pls help

by Chaitanya_1986 » Mon May 02, 2011 10:50 am
17. 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 square regions in terms of r?
1) πr^2
2) πr^2 + 10
3) πr^2 + 1/4π^2r^2
4) πr^2 + (40 - 2πr)^2
5) πr^2 + (10 - 1/2πr)^2

OA is E)
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by manpsingh87 » Mon May 02, 2011 11:00 am
Chaitanya_1986 wrote:17. 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 square regions in terms of r?
1) πr^2
2) πr^2 + 10
3) πr^2 + 1/4π^2r^2
4) πr^2 + (40 - 2πr)^2
5) πr^2 + (10 - 1/2πr)^2

OA is E)
let radius of circle be r, and length of each side of square be x;
therefore 2(pi)r+4x=40;
4x=40-2(pi)r;
x=10-1/2 *(pi)* r;-----1)

total area= area of circle+area of square;
pi*r^2+x^2;
substituting value of x from 1 we have;
pi*r^2+(10-1/2*pi*r);
hence E
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by Chaitanya_1986 » Mon May 02, 2011 11:22 am
Thanks Man....I got it now
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by Tani » Tue May 03, 2011 6:40 am
This would be a great problem to try picking numbers. Let r = 2. The area of the circle is 4pi. , the circumference of the circle is also 4pi.

The remaining wire is 40 - 4pi. That makes each side of the square equal to 10-pi. The area of the square is then (10-pi)^2.

Substituting 2 for r in the answers you see that E = 4pi - (10-pi)^2[/img]
Tani Wolff
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