What's the area of the star /sci/ Wolfram won't help

brainlet here
is it 21.46cm^2?

>>8701069
area of circle / 2

Wish it was harder desu

Just used calculus because it's easier than thinking.

It's trivial that it's symmetrical over each axis. Relevant equation is 5-sqrt(5^2 - (x-5)^2) from 0 to 5. So the answer is
[math]\displaystyle 4 \int_0^5 5-\sqrt{ 5^2 - \left ( x-5 \right ) ^2 } dx[/math] which is also trivial to solve, as you don't really need to evaluate the integral of that root as you can easily get the answer to it.

You end up getting -25(pi-4) which rounds to >>8701116

>>8701121
(also I did use wolfram alpha to evaluate the integral because I didn't want to think :^) )

>>8701069
draw a square around it
inside is just area of square minus area of the circle
get >>8701116

>>8701125
Shit I'm fucking retarded (>>8701121)

>>8701069
draw a square circumscribing the circle. It's clear the outside of the circle is equal to the star shape. Thus the answer is the difference between the area of the square and the circle = 10^2-pi*5^2

No calculus required.

How is this method?

Draw a triangle from the center to the edge. Multiply by 4, subtract from area of circle, multiply by 2.

Impossible to tell without knowing the equations for boundaries of the star.

>>8701069
Exactly [math](100 - 25\pi) cm^2[/math].

100-25pi^2