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Here's a puzzle for when the site activity is low!

 

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)? 

   circles

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 Oct 28, 2015
edited by Alan  Oct 28, 2015

Best Answer 

 #6
avatar+26396 
+35

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

 

circle1:x2+y2=4r1=2circle2:(x2)2+(y2)2=1r2=1

 

intersections S1 and S2:

2x2s2axs+a2r21=0a=3r21r222r1=114S1=(18(117)18(11+7))S2=(18(11+7)18(117))

 

sectors angels (cosinus-rule) α1 and α2:


cos(α1)=5764α1=27.0481105464cos(α2)=916α2=55.7711336722

Areas sectors As1 and As2:

 

As1=4π27.0481105464360As2=π55.7711336722360

Areas triangles At1 and At2:

At1=12|S2×S1|=11327At2=12|[S1(22)]×[S2(22)]|=5327

 

The area of their intersection A=As1At1+As2At2

 

A=π360(427.0481105464+55.7711336722)72A=1.430852126031.32287565553A=0.10797647050

 Oct 28, 2015
 #2
avatar+130458 
+7

I could have solved this with some messy Algebra, but....I decided to cheat....LOL!!!

 

Placing the center of the larger circle at the origin, we have these two equations :

 

x^2 + y^2 = 4      and  (x - 2)^2 + (y - 2)^2 = 1

 

Here's a diagram :

 

 

The area of triangle ABC  =  2*sin(27.048)  = about .909 units^2

 

And the area of the sector ABC = about 944 units^2

 

Similarly, the area of triangle  DAB = (1/2)sin(55.771)  = about  .413  units^2

 

And the area of sector DAB = about .4867 units^2

 

So...the approximate total area  of the  area of intersection = [ .944 - .909] +  [ .4867 - .413] = about .1087 units^2

 

 

cool cool cool

 

OOPS....you are correct Alan....I have provided an edit......duh!!!....I think I mis-placed my decimal point.....Is my answer now correct???

 Oct 28, 2015
edited by CPhill  Oct 28, 2015
 #3
avatar+33654 
+5

Good try Melody, but this shows why it isn't correct:

circ1

Will now have a look at Chris's result (which still doesn't match what I'm expecting!)

 

Ah! If Chris could do adds and take aways correctly he would have the right answer!

 

 

.

 Oct 28, 2015
edited by Alan  Oct 28, 2015
 #4
avatar+118696 
0

Yes I know that Alan - That is why I wrote that it was not correct.

I suppose i should jut have deleted it.  I will delete it now.

 

I am working on a solution that uses co-ordinate geometry.  I am sure I can do it that way but it is quite painful especially if I want to keep the answer exact.  ://

 

There you go Chris, Yours is wrong too.   LOL

But your diagram is fantastic.    laugh

I am only joking Chris,  You answer is great too :))

 

 

 

Thanks for this distraction Alan.  I have enjoyed it.   laugh

 Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
 #5
avatar+33654 
0

I find it fascinating that the question is much more complicated than it appears to be at first sight.

 Oct 28, 2015
 #6
avatar+26396 
+35
Best Answer

Here's a puzzle for when the site activity is low!

In the diagram below there are two quarter circles of radius 1 and 2 intersecting as shown.  What is the area of their intersection (i.e. the shaded region)?

 

circle1:x2+y2=4r1=2circle2:(x2)2+(y2)2=1r2=1

 

intersections S1 and S2:

2x2s2axs+a2r21=0a=3r21r222r1=114S1=(18(117)18(11+7))S2=(18(11+7)18(117))

 

sectors angels (cosinus-rule) α1 and α2:


cos(α1)=5764α1=27.0481105464cos(α2)=916α2=55.7711336722

Areas sectors As1 and As2:

 

As1=4π27.0481105464360As2=π55.7711336722360

Areas triangles At1 and At2:

At1=12|S2×S1|=11327At2=12|[S1(22)]×[S2(22)]|=5327

 

The area of their intersection A=As1At1+As2At2

 

A=π360(427.0481105464+55.7711336722)72A=1.430852126031.32287565553A=0.10797647050

heureka Oct 28, 2015
 #7
avatar+130458 
0

I just had my decimal point in the wrong place....Alan said mine is [fairly] correct....

 

Soooooo.....

 

A BIG GOLD STAR FOR HEUREKA AND ME....!!!!

 

 

 

And for Melody????....A BAG OF SWITCHES....!!!!

 

 

cool cool cool

 Oct 28, 2015
 #8
avatar
0

Area of the square :  2 x 2  = 4

Area of Big arc = 1/4 pi 2^2 = pi

Area of small arc = 1/4 pi 1^2 = 1/4 pi

Area OUTSIDE of Small arc = 4 - 1/4 pi

Area OUTSIDE of Big arc = 4 - pi

 

Add the two outside areas and subtract the square area

 

4-1/4pi + 4 - pi -4 = 4 - 5/4 pi = ,073 Square units

 Oct 28, 2015
 #9
avatar+33654 
0

This is the same as Melody's first answer Guest.  Good try, but unfortunately it isn't correct - see my first reply above to see why.

 Oct 28, 2015
 #10
avatar
0

Guest 8      Ahhhhh......sorry.  I didn't see the other answers before I posted .   I see your explanation of why my answer is WRONG  and it makes perfect sense '2C' it now.......   ha

 Oct 28, 2015

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