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GeezerVison is showing mathworld.wolfram.com — Semiperimeter -- from Wolfram MathWorld

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Semiperimeter

The semiperimeter on a figure is defined as

s=1/2p,
(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

A_Delta=sqrt(s(s-a)(s-b)(s-c)),
(2)

and Brahmagupta's formula for the area of a quadrilateral

A_(quadrilateral)=sqrt((s-a)(s-b)(s-c)(s-d)-abcdcos^2((A+B)/2)).
(3)

The semiperimeter also appears in the beautiful l'Huilier's theorem about spherical triangles.

TriangleSemiperimeter

For a triangle, the following identities hold,

s-a=1/2(-a+b+c)
(4)
s-b=1/2(+a-b+c)
(5)
s-c=1/2(+a+b-c).
(6)

Now consider the above figure. Let I be the incenter of the triangle DeltaABC, 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

BG=BD+AD+AG=BD+AD+CE
(7)
=1/2(2BD+2AD+2CE)
(8)
=1/2[(BD+BE)+(AD+AF)+(CE+CF)]
(9)
=1/2[(BD+AD)+(BE+CE)+(AF+CF)]
(10)
=1/2(AB+BC+AC)
(11)
=1/2(a+b+c)
(12)
=s.
(13)

Furthermore,

s-a=BG-BC
(14)
=(BD+AD+AG)-(BE+CE)
(15)
=(BD+AD+CE)-(BD+CE)
(16)
=AD
(17)
s-b=BG-AC
(18)
=(BD+AD+AG)-(AF+CF)
(19)
=(BD+AD+CE)-(AD+CE)
(20)
=BD
(21)
s-c=BG-AB
(22)
=AG
(23)

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

See also

Perimeter

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References

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.

Referenced on Wolfram|Alpha

Semiperimeter

Cite this as:

Weisstein, Eric W. "Semiperimeter." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/Semiperimeter.html

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