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Math Help - finding the area of the shaded part.

  1. #1
    rcs
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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.
    Attached Thumbnails Attached Thumbnails finding the area of the shaded part.-img_3223.jpg  
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    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
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: finding the area of the shaded part.

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

    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}
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    Re: finding the area of the shaded part.

    The area of a circle of radius 8 is, of course 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 (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 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.
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    rcs
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    Re: finding the area of the shaded part.

    thank you guys
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