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


    Consider the area of one-fifth of the star, quadrilateral $\displaystyle AOBP$

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

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

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

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


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


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


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


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

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

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

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