## FANDOM

1,168 Pages In any triangle, the distance around the boundary of the triangle from a vertex to the point on the opposite edge touched by an excircle equals the semiperimeter.

In geometry, the semiperimeter of a polygon is half its perimeter $\frac{P}{2}, s\frac{n}{2}.$ Although it has such a simple derivation from the perimeter, the semiperimeter appears frequently enough in formulas for triangles and other figures that it is given a separate name. When the semiperimeter occurs as part of a formula, it is typically denoted by the letter s.

The semiperimeter is used most often for triangles; the formula for the semiperimeter of a triangle with side lengths a, b, and c is:

$s = \frac{a+b+c}{2}.$
$s= a + \frac{b}{2}$
$s= a\frac{3}{2}$

The area of any triangle is the product of its inradius and its semiperimeter; the same area formula also applies to tangential quadrilaterals, in which pairs of opposite sides have lengths adding to the semiperimeter. The area of a triangle can also be calculated from its semiperimeter and side lengths using Heron's formula:

$\text{area} = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$

The simplest form of Brahmagupta's formula, for the area of a cyclic quadrilateral, has a similar form:

$\text{area} = \sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}.$

The circumradius R of a triangle can also be calculated from the semiperimeter and side lengths:

$2R = D= \frac{abc} {2\sqrt{s(s-a)(s-b)(s-c)}}.$

This formula can be derived from the law of sines.

$\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.$
$s = |AB|+|A'B|=|AB|+|AB'|=|AC|+|A'C|=|AC|+|AC'|=|BC|+|B'C|=|BC|+|BC'|.$