find the area enclosed by 5 circular arcs
tainangwong shared this question 8 years ago
I want to find the area enclosed by 5 circular arcs but I do not know how to
I would be truly grateful and most thankful if anyone could tell me
If possible, I would like to have the value presented in the simplest exact form
(not the calculated numerical value, instead, keep the surds and fractions)
- Feedback & Questions
- +43 677 6137 2693
I guess the value is likely to be
x π - y
where x and y are non-zero constants
but I am not sure
Hi both, I've tried without solve with another way : use O(0,0) as Center of the polygon(s) and complex analysis : similitary transformations
From A_0(z) (as A) to A_1 (as B) z |--> exp(i 2 pi/5) z
From A_1 to A_0' (as F) using that A_0A_1A_0' is regular so the rotate with centrer AZ_0 an angle pi/3 transforms A_1 to A_0' :
z(A_1) | --> exp(ipi/3)(z(A_1)-z)+z= exp(ipi/3)(exp(i2 pi/5)z-z)+z=coef z
so from poly1 to poly2 we use a z'=coef z so poly2 is regular with the same center.
Note that arg(coef) is -4pi/5.
I can't yeld the exact evaluation of the polygon aera with the CAS even if you can find it : 5/2sin(2pi/5)ray^2
I can't get the exact evaluation of abs(coef) so poly2+5(CircularSector[A_1, A_0', A_1']-Polygon[A_1,A_0',A_1']) does not yeld what i expected
Good luck to find better
Area btw. circle and pentagon side = R^2/2 * (Alpha - SIN(Alpha))
where Alpha represents your "x*pi" and sinus (Alpha) the constant 'y'. Both magnified by the area (R^2/2). Cheers
Nice job Philippe. I wanted to know why this angles about your file : (i note ABC for the value of a geometric angle with center B and from A to C)
regular pentagon : BAN=pi-2pi/5=3pi/5
regular triangle : BAL=pi/3
by reflect : IAL=BAN/2-NAL=3pi/10-4pi/15=pi/30 so KAL=pi/30 so 12°.
For a regular pentagon the area is 5/4 side^2/tan(pi/5)
so poly1=5/4 AB^2/tan(pi/5)
LK=2 AB sin(pi/30) (so, in my last post, the exact value of abs(coef) is 2 cos(pi/30) )
so poly2=5 (AB sin(pi/30))^2/tan(pi/5)
so, if i don't do a mistake, your area is : 5 (AB sin(pi/30))^2/tan(pi/5)+5AB^2/2(pi/15-sin(pi/15))=5 AB^2((sin(pi/30))^2/tan(pi/5)+1/2(pi/15-sin(pi/15)))
5 AB^2((sin(pi/30))^2/tan(pi/5)+1/30 pi-1/2 sin(pi/15))
What I've done, up to now (tomorrow or later I'll see your last worksheet), is to point out some parameters governing the relationships between both pentagons. So, I realized that the piece of area between the circle of generic radius R can be quickly computed by subtracting the sector area of amplitude Alpha and the subtended triangle, as specified. This matches what the user (post opener) was expected.
Of course, different solutions are possible using mathematics, better than dynamic geometry like GG which generally covers numerical analysis, not ... brained investigations. Thx&Cheers
Back to you, Michel, to show our coincident values. Cheers
thank you !!!
Comments have been locked on this page!