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**hussam****Member**- Registered: 2015-04-06
- Posts: 46

hi,

you have a circle in a squere , the blue area is included in the circle

like this Mr bobbym?

*Last edited by hussam (2015-04-09 07:02:18)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Are they saying that the area of the blue region is

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**hussam****Member**- Registered: 2015-04-06
- Posts: 46

hi ;

exactly

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

You just need to manipulate this identity where the LHS is the area of the grey region and the RHS is the area of the pink.

where s is the side of the square?

I think the problem should provide a bit more information because many assumptions had to be made. For one thing, is the polygon a square? Is that circle inscribed in the square? I have deduced those from the drawing but it would be much better if the problem stated it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**hussam****Member**- Registered: 2015-04-06
- Posts: 46

a simple method :

the gray area is 3( r^2 - 1/4 π r^2)

the pink area is 1/4π r^2 -(π-3) r^2

= r^2(1/4π-π+3)

=r^2(-3/4π+3)

=3 r^2(1-1/4π)

=3(r^2-1/4π r^2)=the gray area

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