- geometry, the
semiperimeter of a
polygon is half its perimeter.
Although it has such a
simple derivation from the perimeter, the
semiperimeter appears frequently...
- {\displaystyle b,} c . {\displaystyle c.}
Letting s {\displaystyle s} be the
semiperimeter of the triangle, s = 1 2 ( a + b + c ) , {\displaystyle s={\tfrac {1}{2}}(a+b+c)...
- {\displaystyle \triangle ABC} with
sides a ≤ b < c {\displaystyle a\leq b<c} ,
semiperimeter s = 1 2 ( a + b + c ) {\textstyle s={\tfrac {1}{2}}(a+b+c)} , area T...
-
sides a {\displaystyle a} , b {\displaystyle b} , c {\displaystyle c} ,
semiperimeter s {\displaystyle s} , area T {\displaystyle T} ,
exradii r a {\displaystyle...
-
perimeter into two
equal lengths, this
common length being called the
semiperimeter of the triangle. The
three splitters of a
triangle all
intersect each...
- ( a + b + c + d ) {\displaystyle s={\tfrac {1}{2}}(a+b+c+d)} is the
semiperimeter of the trapezoid. (This
formula is
similar to Brahmagupta's formula...
- = 1 2 ( a + b + c ) {\displaystyle s={\tfrac {1}{2}}(a+b+c)} is the
semiperimeter. The
tangency points of the
incircle divide the
sides into segments...
- (orig. 1960). Gutierrez, Antonio, "Hexagon,
Inscribed Circle, Tangent,
Semiperimeter", [1]
Archived 2012-05-11 at the
Wayback Machine,
Accessed 2012-04-17...
- − d ) {\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}
where s, the
semiperimeter, is
defined to be s = a + b + c + d 2 . {\displaystyle s={\frac {a+b+c+d}{2}}...
-
inequality holding for an
obtuse triangle. For an
acute triangle with
semiperimeter s,: p.115, #2874 s − r > 2 R , {\displaystyle s-r>2R,} and the reverse...
