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:(x−2)2+(y−2)2=1r2=1
intersections S1 and S2:
2x2s−2a⋅xs+a2−r21=0a=3r21−r222r1=114→S1=(18(11−√7)18(11+√7))→S2=(18(11+√7)18(11−√7))
sectors angels (cosinus-rule) α1 and α2:
cos(α1)=5764α1=27.0481105464∘cos(α2)=916α2=55.7711336722∘
Areas sectors As1 and As2:
As1=4π27.0481105464∘360∘As2=π55.7711336722∘360∘
Areas triangles At1 and At2:
At1=12⋅|→S2×→S1|=1132√7At2=12⋅|[→S1−(22)]×[→S2−(22)]|=532√7
The area of their intersection A=As1−At1+As2−At2
A=π360(4⋅27.0481105464∘+55.7711336722∘)−√72A=1.43085212603−1.32287565553A=0.10797647050
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
OOPS....you are correct Alan....I have provided an edit......duh!!!....I think I mis-placed my decimal point.....Is my answer now correct???
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.
I am only joking Chris, You answer is great too :))
Thanks for this distraction Alan. I have enjoyed it.
I find it fascinating that the question is much more complicated than it appears to be at first sight.
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:(x−2)2+(y−2)2=1r2=1
intersections S1 and S2:
2x2s−2a⋅xs+a2−r21=0a=3r21−r222r1=114→S1=(18(11−√7)18(11+√7))→S2=(18(11+√7)18(11−√7))
sectors angels (cosinus-rule) α1 and α2:
cos(α1)=5764α1=27.0481105464∘cos(α2)=916α2=55.7711336722∘
Areas sectors As1 and As2:
As1=4π27.0481105464∘360∘As2=π55.7711336722∘360∘
Areas triangles At1 and At2:
At1=12⋅|→S2×→S1|=1132√7At2=12⋅|[→S1−(22)]×[→S2−(22)]|=532√7
The area of their intersection A=As1−At1+As2−At2
A=π360(4⋅27.0481105464∘+55.7711336722∘)−√72A=1.43085212603−1.32287565553A=0.10797647050
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....!!!!
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