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Math Help - Segments formed by an inscribed equilateral triangle

  1. #1
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    Segments formed by an inscribed equilateral triangle

    Hello. This is a problem from a book I'm studying on, I can't understand the solution.

    "Find each segment formed by an inscribed equilateral triangle if the radius of the circle is 8"

    Solution
    R=8. Since s=R\sqrt{3}=8\sqrt{3}, the area of \vartriangle ABC is \tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}.
    Also, area of circle O = \pi R^{2}=64\pi .
    Hence area of segment BDC = \tfrac{1}{3}\left( 64\pi -48\sqrt{3} \right)

    These are my questions:
    1.- Why is s=R\sqrt{3}=8\sqrt{3}?

    2.And why is the area of \vartriangle ABC equal to \tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}? I know the area of a triangle is equal to \tfrac{1}{2}bh, how do you get the height?
    Thanks.
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  2. #2
    A riddle wrapped in an enigma
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    Inscribed equilateral triangle:

    Draw apothem from center perpendicular to a side.

    Draw radius from center to vertex of same side.

    Triangle formed is 30-60-90.

    If hypotenuse(radius) is 8, then side opposite the 30 deg. angle is 4 (apothem).

    Side opposite 60 degree angle is 4\sqrt3

    Thus, the length of a side is 2(4\sqrt3)=8\sqrt3

    Perimeter = 3(8\sqrt3)=24\sqrt3

    Area=\frac{1}{2}Pa, where P=perimeter, a=apothem

    Finally, A=\frac{1}{2}(24\sqrt3)(4)=48\sqrt3
    Attached Thumbnails Attached Thumbnails Segments formed by an inscribed equilateral triangle-circle.jpg  
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  3. #3
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    Thanks a lot.
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