Find the Area of the shaded region in the given problem

  • Thread starter Thread starter chwala
  • Start date Start date
  • Tags Tags
    Area
AI Thread Summary
The discussion revolves around finding the area of a shaded region in a geometry problem, with one participant expressing initial confusion but ultimately arriving at a solution. They reference a textbook answer of 17.5 and plan to double-check their rounding. The participant shares their working formula, which involves calculating areas using trigonometric functions and geometric principles. They seek confirmation on whether their approach aligns with others in the discussion. The conversation highlights the complexity of the problem and the collaborative effort to verify solutions.
chwala
Gold Member
Messages
2,827
Reaction score
415
Homework Statement
See attached
Relevant Equations
Circular measure
Wawawawawa boggled me a little bit... but finally managed it...seeking alternative approach guys;

kindly note that what i have indicated as ##*## and a ##√## is the correct working ...

Text book answer indicates ##17.5## as answer... will re check my rounding solutions later...
1678364583516.png


My working- allow me to copy paste here...will later type the working...

1678364659756.png

1678364699766.png
 
Last edited:
Physics news on Phys.org
From your sketch, say S is area we want, it seems
2S=\frac{1}{2}*20^2 (4\alpha)-4*\frac{1}{2}*15\cos \alpha*5-2*\frac{1}{2}*5^2(\pi+2\alpha)
where
\sin\alpha =\frac{1}{3}
Is it same as your idea ?
 
Last edited:
  • Like
Likes neilparker62, chwala and Steve4Physics
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top