Exponents and roots are what operations
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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°.
The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution