The routine area returns the area of the input object which is a triangle, a square, a circle, or an ellipse.

The command with(geometry,area) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{geometry}\right)\:$

$\mathrm{triangle}\left(\mathrm{ABC}\,\left[\mathrm{point}\left(A\,0\,0\right)\,\mathrm{point}\left(B\,2\,0\right)\,\mathrm{point}\left(C\,1\,3\right)\right]\right)\:$

$\mathrm{area}\left(\mathrm{ABC}\right)$

$3$

$\mathrm{square}\left(\mathrm{Sqr}\,\left[\mathrm{point}\left(A\,\left[1\,-2^{\frac{1}{2}}+1\right]\right)\,\mathrm{point}\left(B\,\left[2^{\frac{1}{2}}+1\,1\right]\right)\,\mathrm{point}\left(C\,\left[1\,2^{\frac{1}{2}}+1\right]\right)\,\mathrm{point}\left(E\,\left[-2^{\frac{1}{2}}+1\,1\right]\right)\right]\right)\:$

$\mathrm{area}\left(\mathrm{Sqr}\right)$

$4$

$\mathrm{circle}\left(c\,\left[\mathrm{point}\left(\mathrm{OO}\,-3\,2\right)\,2\right]\right)\:$

$\mathrm{area}\left(c\right)$

$4\mathrm{\pi }$

$\mathrm{ellipse}\left(e\,\frac{x^2}{16}+\frac{y^2}{9}=1\,\left[x\,y\right]\right)\:$

$\mathrm{area}\left(e\right)$

$12\mathrm{\pi }$