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finding the area of the shaded part.

there some parts of the solution on the book where i found this problem.

Area of shaded region = A sector + [ A triangle - 2 (A sector 2)]

= 5/6 A circle + [ (s^2 * sqrt of 3)/4 - 2(1/6 A of circle)]

... i understand the 5/6 A circle and the 2(1/6 Area of circle, but i dont understand the (s^2*sqrt of 3)/4 for the Area of triangle.... how was that being done? please help me understand this. Thank you.

Re: finding the area of the shaded part.

I am not looking at what extra info you provide but try this

area 8 radius circle +equilateral triangle side lenght 8 - 3 times the segment of a sector of 8 radius where sector is 1/6 of circle

Re: finding the area of the shaded part.

The area $\displaystyle A$ of an equilateral triangle having sides $\displaystyle s$ may be computed as follows:

$\displaystyle A=\frac{1}{2}\cdot s\cdot s\cdot\sin(60^{\circ})=\frac{1}{2}s^2\cdot\frac{ \sqrt{3}}{2}=\frac{s^2\sqrt{3}}{4}$

Re: finding the area of the shaded part.

The area of a circle of radius 8 is, of course $\displaystyle 64\pi$. Every angle of an equilateral triangle has measure 60 degrees which is 60/360= 1/6 of the entire circle. The portion of the circle **outside** the triangle has area $\displaystyle (5/6)(64\pi)$. If you draw a horizontal line where the triangle cuts the circle, you have an equilateral triangle with side length 8 and can use the formulas given above.

The area of the trapezoid below that has bases 8 and 16 and height $\displaystyle 4\sqrt{3}$ and the the area of such a trapezoid is (h/2)(b1+ b2). Now you have to **remove** unshaded area in that trapezoid which is again the area of 1/6 of a circle of radius 8.

Re: finding the area of the shaded part.