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
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: .$\displaystyle \text{Area of star} \:=\:\tfrac{5}{2}\left(R^2\sin72^o - a^2\sin108^o\right)$
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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]: .$\displaystyle 2a + 2a\sin18^o \:=\:2R\sin72^o$
. . $\displaystyle a(1+\sin18^o) \:=\:R\sin72^o \quad\Rightarrow\quad a \:=\:\frac{R\sin72^o}{1+\sin18^o}$