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Math Help - Finding area of star in circle.

  1. #1
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    Finding area of star in circle.

    Hi i need to find the area of the star only knowing the radius of the circle: r=1


    For everything thanks in advance
    Last edited by mr fantastic; November 27th 2011 at 10:23 AM. Reason: Re-titled.
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  2. #2
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    Re: Can someone help me to find this area?

    Quote Originally Posted by leart369 View Post
    Hi i need to find the area of the star only knowing the radius of the circle: r=1

    For everything thanks in advance
    Have a look here: Pentagram - Wikipedia, the free encyclopedia
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  3. #3
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    Re: Can someone help me to find this area?

    Quote Originally Posted by earboth View Post
    I don't get it
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  4. #4
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    Re: Can someone help me to find this area?

    Hello, leart369!

    I have a very "clunky" solution . . .


    i need to find the area of the star only knowing the radius of the circle: R=1

    Label the vertices A,B,C,D,E starting at the top and moving clockwise.
    . . Draw segments AB,BC,CD,DE,EA.
    Let O be the center.
    \text{Let }P\,=\,AC \cap BE.\;\text{Let }Q \,=\,AD \cap BE.


    Consider the area of one-fifth of the star, quadrilateral AOBP

    The area of a triangle is given by: . A \:=\:\tfrac{1}{2}ab\sin C
    . . (one-half the product of two sides and the sine of the included angle)

    The area of \Delta AOB \:=\:\tfrac{1}{2}R^2\sin72^o

    The area of \Delta APB \:=\:\tfrac{1}{2}a^2\sin108^o

    . . The area of quad AOBP \:=\:\tfrac{1}{2}R^2\sin72^o - \tfrac{1}{2}a^2\sin108^o


    Therefore: . \text{Area of star} \:=\:\tfrac{5}{2}\left(R^2\sin72^o - a^2\sin108^o\right)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We know that R = 1. . Can we find a\,? . . . . Yes!


    \begin{array}{cccccc}\text{In }\Delta APQ\!: & b \:=\:2a\sin18^o & [1] \\ \text{In }\Delta AOC\!: & 2a+b \:=\:2R\sin72^o & [2] \end{array}

    Substitute [1] into [2]: . 2a + 2a\sin18^o \:=\:2R\sin72^o

    . . a(1+\sin18^o) \:=\:R\sin72^o \quad\Rightarrow\quad a \:=\:\frac{R\sin72^o}{1+\sin18^o}

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