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For a convex cyclic quadrilateral with successive side lengths $a,b,c,d$ and semiperimeter $s=(a+b+c+d)/2$, Brahmagupta's formula gives its area:The cyclic hypothesis is essential. To derive it, let opposite interior angles be $B,D$, so $B+D=\pi$. Splitting along their common opposite diagonal of a polygon gives $2K=(ab+cd)\sin B$, while the law of cosines gives $2(ab+cd)\cos B=a^2+b^2-c^2-d^2$. Eliminate $\cos B$ using ${\sin }^{2}B+{\cos }^{2}B=1$ and factor the result to obtain $16K^2=16(s-a)(s-b)(s-c)(s-d)$. For a rectangle with sides $a,b,a,b$, the formula reduces to $K=ab$.

For a triangle with side lengths $a,b,c$ and semiperimeter $s=(a+b+c)/2$, its area is $A=\sqrt{s(s-a)(s-b)(s-c)}$. This expresses the area using only the three side lengths.

The inradius is the radius of an incircle. For a triangle of area $A$ and semiperimeter $s$, it is $r=A/s$: joining the incenter to the three vertices partitions the triangle into three triangles with common altitude $r$.