Your teacher is giving you
1 answer:
You might be interested in
Leeza can put 33 or less gallons of water in aquarium.
Step-by-step explanation:
Given,
Capacity of aquarium = 55 gallons of water
Water already in aquarium = 22 gallons
Let,
x be the number of gallons that can be added in aquarium.
As aquarium can hold 55 gallons of water, therefore, we cannot add more than 55 gallons of water.
Subtracting 22 from both sides
Leeza can put 33 or less gallons of water in aquarium.
Keywords: inequality, subtraction
Learn more about inequalities at:
#LearnwithBrainly
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°.