# 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)

[size=150]Great thanks!!![/size]

1

I guess the value is likely to be

x π - y

where x and y are non-zero constants

but I am not sure

1

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

Cheers. Michel

https://ggbm.at/569425

1

I guess the value is likely to be

x π - y

where x and y are non-zero constants

but I am not sure

It is so! In fact, according to the attachments (screenshot+GG-sheet), the separation angle of the tiny pentagon side seen from the opposite vertex of the big one is 1/9 of the internal angle, so:

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

Philippe

Files: 19.png
1

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

NAL=3pi/5-pi/3=4pi/15

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)

Area btw. circle and pentagon side = R^2/2 * (Alpha - SIN(Alpha))

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))

Cheers. Michel

https://ggbm.at/569439

1

Hi Michel!

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

Philippe

1

Back to you, Michel, to show our coincident values. Cheers

Philippe

Files: 21.png
1

brilliant !!!

thank you !!!