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The semiperimeter on a figure is defined as

s\equiv {\textstyle{1\over 2}}p,
\end{displaymath} (1)

where $p$ is the Perimeter. The semiperimeter of Polygons appears in unexpected ways in the computation of their Areas. The most notable cases are in the Altitude, Exradius, and Inradius of a Triangle, the Soddy Circles, Heron's Formula for the Area of a Triangle in terms of the legs $a$, $b$, and $c$
\end{displaymath} (2)

and Brahmagupta's Formula for the Area of a Quadrilateral

A_{\rm quadrilateral}=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left({A+B\over 2}\right)}.
\end{displaymath} (3)

The semiperimeter also appears in the beautiful L'Huilier's Theorem about Spherical Triangles.


For a Triangle, the following identities hold,

$\displaystyle s-a$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(-a+b+c)$ (4)
$\displaystyle s-b$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(a+b-c)$ (5)
$\displaystyle s-c$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(a+b-c).$ (6)

Now consider the above figure. Let $I$ be the Incenter of the Triangle $\Delta ABC$, with $D$, $E$, and $F$ the tangent points of the Incircle. Extend the line $BA$ with $GA=CE$. Note that the pairs of triangles $(ADI, AFI)$, $(BDI, BEI)$, $(CFI, CEI)$ are congruent. Then
$\displaystyle BG$ $\textstyle =$ $\displaystyle BD+AD+AG=BD+AD+CE$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(2BD+2AD+2CE)$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}[(BD+BE)+(AD+AF)+(CE+CF)]$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}[(BD+AD)+(BE+CE)+(AF+CF)]$  
  $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}(AB+BC+AC)={\textstyle{1\over 2}}(a+b+c)=s.$ (7)

$\displaystyle s-a$ $\textstyle =$ $\displaystyle BG-BC$  
  $\textstyle =$ $\displaystyle (BD+AD+AG)-(BE+CE)$  
  $\textstyle =$ $\displaystyle (BD+AD+CE)-(BD+CE)=AC$ (8)
$\displaystyle s-b$ $\textstyle =$ $\displaystyle BG-AC$  
  $\textstyle =$ $\displaystyle (BD+AD+AG)-(AF+CF)$  
  $\textstyle =$ $\displaystyle (BD+AD+CE)-(AD+CE)=BD$ (9)
$\displaystyle s-c$ $\textstyle =$ $\displaystyle BG-AB=AG$ (10)

(Dunham 1990). These equations are some of the building blocks of Heron's derivation of Heron's Formula.


Dunham, W. ``Heron's Formula for Triangular Area.'' Ch. 5 in Journey Through Genius: The Great Theorems of Mathematics. New York: Wiley, pp. 113-132, 1990.

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© 1996-9 Eric W. Weisstein