This is Area of composite figures
1 answer:
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Answer: So, the probability will be
Step-by-step explanation:
Since we have given that
Number of blue balls =30
Number of red balls = 40
Number of white balls = 30
Since there are 100 balls in total ,
As we have given that there is no white ball chosen.
So, only 70 balls are left with us , from which we choose a blue ball
So, the probability will be
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.
![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)
![\bf A_y=\cfrac{6^2}{2}\left[\cfrac{\pi\cdot 120 }{180}~-~sin(120^o ) \right] \\\\\\ A_y=18\left[\cfrac{2\pi }{3}~-~\cfrac{\sqrt{3}}{2} \right]\implies \boxed{A_y=12\pi -9\sqrt{3}}\\\\ -------------------------------\\\\ \textit{shaded area}\qquad \stackrel{A_x}{6\pi }~~+~~\stackrel{A_y}{12\pi -9\sqrt{3}}\implies 18\pi -9\sqrt{3}](https://tex.z-dn.net/?f=%5Cbf%20A_y%3D%5Ccfrac%7B6%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%5Ccdot%20120%20%7D%7B180%7D~-~sin%28120%5Eo%20%29%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0AA_y%3D18%5Cleft%5B%5Ccfrac%7B2%5Cpi%20%7D%7B3%7D~-~%5Ccfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%20%5Cright%5D%5Cimplies%20%5Cboxed%7BA_y%3D12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bshaded%20area%7D%5Cqquad%20%5Cstackrel%7BA_x%7D%7B6%5Cpi%20%7D~~%2B~~%5Cstackrel%7BA_y%7D%7B12%5Cpi%20-9%5Csqrt%7B3%7D%7D%5Cimplies%2018%5Cpi%20-9%5Csqrt%7B3%7D)
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.