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perimeter ?

a circle whose area is 2/pie.a rectangle inside the circle.then find the parameter of that rectangle.

Note by Google Face
4 years, 7 months ago

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Comment deleted Apr 27, 2013

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k doesn't matter ...

Google Face - 4 years, 7 months ago

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I think 8/π is right answer either it is square or rectangle. because in both case both diagonal will be same, because it is diameter of circle. So even by changing length and width of the rectangle parameter will be same in all case. and that can be find by assuming that rectangle as a square..

Bharat Bhawsar - 3 years, 9 months ago

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its 8/pie.....i calculated so..

Anurag Nayan - 4 years, 7 months ago

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how please explain

Google Face - 4 years, 7 months ago

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I took the rectangle as a square. If that's not the case, then the problem is lacking some information i guess.

If we consider the rectangle to be a square, then here goes the solution: We have, \(\large \pi\times r^2=\large \frac{2}{\pi} \implies \boxed {r=\frac{\sqrt{2}}{\pi}} \) Now, let the side of square be \(x\). Then \(\large 2\times x^2=\large \frac{8}{\pi^2} \implies x=\large \frac{2}{\pi}\) This means the perimeter is \(\large 4\times \large \frac{2}{\pi}=\large \boxed {\boxed{\frac{8}{\pi}}} \)

Vikram Waradpande - 4 years, 7 months ago

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yeah \(\Large \frac{8}{\pi}\) is correct

Vikram Waradpande - 4 years, 7 months ago

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I think many answers are possible. It's 8/π only when the rectangle is a square, as in a rectangle other than the square the angle made by the two radii by either the length or breadth is not 90 degrees so we can't apply pythagoras theorem here.

Bhargav Das - 4 years, 7 months ago

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@Bhargav Das Yeah I took the rectangle to be a square.

Vikram Waradpande - 4 years, 7 months ago

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