+0  
 
+8
784
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avatar+26328 

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

.

Alan  Oct 28, 2015
edited by Alan  Oct 28, 2015

Best Answer 

 #6
avatar+18712 
+33

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)?

 

\(circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\\)

 

intersections \(S_1 \text{ and } S_2\):

\(2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }\)

 

sectors angels (cosinus-rule) \(\alpha_1 \text{ and } \alpha_2\):


\( \cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\\)

Areas sectors \(A_{s_1} \text{ and } A_{s_2}\):

 

\( A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\\)

Areas triangles \(A_{t_1} \text{ and } A_{t_2}\):

\( A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7} \)

 

\( \text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}\)

 

\(A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}\)

heureka  Oct 28, 2015
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9+0 Answers

 #2
avatar+78577 
+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???

CPhill  Oct 28, 2015
edited by CPhill  Oct 28, 2015
 #3
avatar+26328 
+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!

 

 

.

Alan  Oct 28, 2015
edited by Alan  Oct 28, 2015
 #4
avatar+90996 
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

Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
edited by Melody  Oct 28, 2015
 #5
avatar+26328 
0

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

Alan  Oct 28, 2015
 #6
avatar+18712 
+33
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)?

 

\(circle_1: x^2 + y^2 = 4 \qquad r_1 = 2\\ circle_2: (x-2)^2 + (y-2)^2 = 1 \qquad r_2 = 1\\\)

 

intersections \(S_1 \text{ and } S_2\):

\(2 x_{s}^2 - 2a \cdot x_s + a^2 - r_1^2 = 0 \qquad a = \frac{3r_1^2-r_2^2}{2r_1} = \frac{11}{4}\\ \vec{S_1} =\dbinom{ \frac18(11-\sqrt{7}) }{ \frac18(11+\sqrt{7}) }\\ \vec{S_2} =\dbinom{ \frac18(11+\sqrt{7}) }{ \frac18(11-\sqrt{7}) }\)

 

sectors angels (cosinus-rule) \(\alpha_1 \text{ and } \alpha_2\):


\( \cos{(\alpha_1)} = \frac{57}{64} \qquad \alpha_1 = 27.0481105464^{\circ}\\ \cos{(\alpha_2)} = \frac{9}{16} \qquad \alpha_2 = 55.7711336722^{\circ}\\\)

Areas sectors \(A_{s_1} \text{ and } A_{s_2}\):

 

\( A_{s_1} = 4\pi \frac{27.0481105464^{\circ}}{360^{\circ}}\\ A_{s_2} = \pi \frac{55.7711336722^{\circ}}{360^{\circ}}\\\)

Areas triangles \(A_{t_1} \text{ and } A_{t_2}\):

\( A_{t_1} = \frac{1}{2} \cdot \left| \vec{S_2}\times \vec{S_1} \right| = \frac{11}{32}\sqrt{7}\\ A_{t_2} = \frac{1}{2} \cdot \left| \left[\vec{S_1}-\binom{2}{2}\right] \times \left[\vec{S_2}-\binom{2}{2}\right] \right|= \frac{5}{32}\sqrt{7} \)

 

\( \text{The area of their intersection }\quad A = A_{s_1}-A_{t_1}+A_{s_2}-A_{t_2}\)

 

\(A=\frac{\pi}{360}(4\cdot 27.0481105464^{\circ}+55.7711336722^{\circ})-\frac{ \sqrt{7} }{2}\\ A= 1.43085212603-1.32287565553\\ \mathbf{A=0.10797647050}\)

heureka  Oct 28, 2015
 #7
avatar+78577 
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

CPhill  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

Guest Oct 28, 2015
 #9
avatar+26328 
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.

Alan  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

Guest Oct 28, 2015

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