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#1 2015-04-09 05:24:39
- hussam
- Member
- Registered: 2015-04-06
- Posts: 46
circle with image (;
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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#2 2015-04-09 05:36:52
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: circle with image (;
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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#3 2015-04-09 05:38:03
- hussam
- Member
- Registered: 2015-04-06
- Posts: 46
Re: circle with image (;
hi ;
exactly
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#4 2015-04-09 05:45:34
- bobbym
- bumpkin
- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606
Re: circle with image (;
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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#5 2015-04-09 06:23:01
- hussam
- Member
- Registered: 2015-04-06
- Posts: 46
Re: circle with image (;
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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