Un triángulo con lados a , b , c e ángulos opostos
α
,
β
,
γ
{\displaystyle \alpha ,\beta ,\gamma }
En xeometría , a fórmula de Herón dá a área dun triángulo en función das tres lonxitudes dos lados
a
,
{\displaystyle a,}
b
,
{\displaystyle b,}
c
.
{\displaystyle c.}
s
{\displaystyle s}
semiperímetro do triángulo,
s
=
1
2
(
a
+
b
+
c
)
,
{\displaystyle s={\tfrac {1}{2}}(a+b+c),}
A
{\displaystyle A}
[ 1]
A
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}.}
Leva o nome do enxeñeiro Herón de Alexandría do século I quen a demostrou na súa obra Metrica , aínda que probablemente se coñecía séculos antes.
Sexa
△
A
B
C
{\displaystyle \triangle ABC}
a
=
4
,
{\displaystyle a=4,}
b
=
13
,
{\displaystyle b=13,}
c
=
15.
{\displaystyle c=15.}
s
=
1
2
(
a
+
b
+
c
)
=
{\displaystyle s={\tfrac {1}{2}}(a+b+c)={}}
1
2
(
4
+
13
+
15
)
=
16
{\displaystyle {\tfrac {1}{2}}(4+13+15)=16}
A
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
16
⋅
(
16
−
4
)
⋅
(
16
−
13
)
⋅
(
16
−
15
)
=
16
⋅
12
⋅
3
⋅
1
=
576
=
24.
{\displaystyle {\begin{aligned}A&={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}\\&={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.\end{aligned}}}
A fórmula de Herón tamén se pode escribir en termos só das lonxitudes, sen usar o semiperímetro,
A
=
1
4
(
a
+
b
+
c
)
(
−
a
+
b
+
c
)
(
a
−
b
+
c
)
(
a
+
b
−
c
)
.
A
=
1
4
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
.
{\displaystyle {\begin{aligned}A&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}.\\[6mu]A&={\tfrac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.\end{aligned}}}
A mesma relación pódese expresar usando o determinante de Cayley-Menger , [ 2]
−
16
A
2
=
|
0
a
2
b
2
1
a
2
0
c
2
1
b
2
c
2
0
1
1
1
1
0
|
.
{\displaystyle -16A^{2}={\begin{vmatrix}0&a^{2}&b^{2}&1\\a^{2}&0&c^{2}&1\\b^{2}&c^{2}&0&1\\1&1&1&0\end{vmatrix}}.}
Proba trigonométrica mediante a lei dos cosenos [ editar | editar a fonte ] En primeiro lugar mostramos unha proba moderna, bastante diferente da proporcionada por Herón. [ 3]
a
,
{\displaystyle a,}
b
,
{\displaystyle b,}
c
{\displaystyle c}
α
,
{\displaystyle \alpha ,}
β
,
{\displaystyle \beta ,}
γ
{\displaystyle \gamma }
ángulos opostos a eses lados (figura inicial). Aplicando a lei dos cosenos obtemos
cos
γ
=
a
2
+
b
2
−
c
2
2
a
b
{\displaystyle \cos \gamma ={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}
A partir deste coseno obtemos o seno,
sin
γ
=
1
−
cos
2
γ
=
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
2
a
b
.
{\displaystyle \sin \gamma ={\sqrt {1-\cos ^{2}\gamma }}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.}
A altura do triángulo con base
a
{\displaystyle a}
b
sin
γ
{\displaystyle b\sin \gamma }
A
=
1
2
(
base
)
(
altura
)
=
1
2
a
b
sin
γ
=
a
b
4
a
b
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
=
1
4
−
a
4
−
b
4
−
c
4
+
2
a
2
b
2
+
2
a
2
c
2
+
2
b
2
c
2
=
1
4
(
a
+
b
+
c
)
(
−
a
+
b
+
c
)
(
a
−
b
+
c
)
(
a
+
b
−
c
)
=
(
a
+
b
+
c
2
)
(
−
a
+
b
+
c
2
)
(
a
−
b
+
c
2
)
(
a
+
b
−
c
2
)
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle {\begin{aligned}A&={\tfrac {1}{2}}({\mbox{base}})({\mbox{altura}})\\[6mu]&={\tfrac {1}{2}}ab\sin \gamma \\[6mu]&={\frac {ab}{4ab}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[6mu]&={\sqrt {\left({\frac {a+b+c}{2}}\right)\left({\frac {-a+b+c}{2}}\right)\left({\frac {a-b+c}{2}}\right)\left({\frac {a+b-c}{2}}\right)}}\\[6mu]&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}}}
Proba alxébrica mediante o teorema de Pitágoras [ editar | editar a fonte ] Triángulo coa altura h que corta a base c en d + (c − d ) A seguinte proba é moi semellante á dada por Raifaizen. [ 4] teorema de Pitágoras temos
b
2
=
h
2
+
d
2
{\displaystyle b^{2}=h^{2}+d^{2}}
a
2
=
h
2
+
(
c
−
d
)
2
{\displaystyle a^{2}=h^{2}+(c-d)^{2}}
a
2
−
b
2
=
c
2
−
2
c
d
.
{\displaystyle a^{2}-b^{2}=c^{2}-2cd.}
d
{\displaystyle d}
d
=
−
a
2
+
b
2
+
c
2
2
c
.
{\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}.}
Para a altura do triángulo temos
h
2
=
b
2
−
d
2
.
{\displaystyle h^{2}=b^{2}-d^{2}.}
d
{\displaystyle d}
diferenza de cadrados obtemos
h
2
=
b
2
−
(
−
a
2
+
b
2
+
c
2
2
c
)
2
=
(
2
b
c
−
a
2
+
b
2
+
c
2
)
(
2
b
c
+
a
2
−
b
2
−
c
2
)
4
c
2
=
(
(
b
+
c
)
2
−
a
2
)
(
a
2
−
(
b
−
c
)
2
)
4
c
2
=
(
b
+
c
−
a
)
(
b
+
c
+
a
)
(
a
+
b
−
c
)
(
a
−
b
+
c
)
4
c
2
=
2
(
s
−
a
)
⋅
2
s
⋅
2
(
s
−
c
)
⋅
2
(
s
−
b
)
4
c
2
=
4
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
c
2
.
{\displaystyle {\begin{aligned}h^{2}&=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}\\&={\frac {(2bc-a^{2}+b^{2}+c^{2})(2bc+a^{2}-b^{2}-c^{2})}{4c^{2}}}\\&={\frac {{\big (}(b+c)^{2}-a^{2}{\big )}{\big (}a^{2}-(b-c)^{2}{\big )}}{4c^{2}}}\\&={\frac {(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^{2}}}\\&={\frac {2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^{2}}}\\&={\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}.\end{aligned}}}
E por último:
A
=
c
h
2
=
c
2
4
⋅
4
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
c
2
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle {\begin{aligned}A&={\frac {ch}{2}}\\&={\sqrt {{\frac {c^{2}}{4}}\cdot {\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}}}
Proba trigonométrica mediante a lei das cotanxentes [ editar | editar a fonte ] Significado xeométrico de s − a s − b s − c lei das cotanxentes para coñecer o razoamento detrás disto. Se
r
{\displaystyle r}
círculo inscrito do triángulo, entón o triángulo pódese dividir en tres triángulos de igual altitude
r
{\displaystyle r}
a
,
{\displaystyle a,}
b
,
{\displaystyle b,}
c
.
{\displaystyle c.}
A
=
1
2
a
r
+
1
2
b
r
+
1
2
c
r
=
r
s
,
{\displaystyle A={\tfrac {1}{2}}ar+{\tfrac {1}{2}}br+{\tfrac {1}{2}}cr=rs,}
onde
s
=
1
2
(
a
+
b
+
c
)
{\displaystyle s={\tfrac {1}{2}}(a+b+c)}
O triángulo pódese dividir alternativamente en seis triángulos (en pares congruentes) de altura
r
{\displaystyle r}
s
−
a
,
{\displaystyle s-a,}
s
−
b
,
{\displaystyle s-b,}
s
−
c
{\displaystyle s-c}
lei das cotanxentes )
A
=
r
(
s
−
a
)
+
r
(
s
−
b
)
+
r
(
s
−
c
)
=
r
2
(
s
−
a
r
+
s
−
b
r
+
s
−
c
r
)
=
r
2
(
cot
α
2
+
cot
β
2
+
cot
γ
2
)
=
r
2
(
cot
α
2
cot
β
2
cot
γ
2
)
=
r
2
(
s
−
a
r
⋅
s
−
b
r
⋅
s
−
c
r
)
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
r
.
{\displaystyle {\begin{aligned}A&=r(s-a)+r(s-b)+r(s-c)\\[2mu]&=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\[2mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\[3mu]&={\frac {(s-a)(s-b)(s-c)}{r}}.\end{aligned}}}
O paso medio anterior é
cot
α
2
+
cot
β
2
+
cot
γ
2
=
{\textstyle \cot {\tfrac {\alpha }{2}}+\cot {\tfrac {\beta }{2}}+\cot {\tfrac {\gamma }{2}}={}}
cot
α
2
cot
β
2
cot
γ
2
,
{\displaystyle \cot {\tfrac {\alpha }{2}}\cot {\tfrac {\beta }{2}}\cot {\tfrac {\gamma }{2}},}
α
2
+
β
2
+
γ
2
=
π
2
.
{\textstyle {\tfrac {\alpha }{2}}+{\tfrac {\beta }{2}}+{\tfrac {\gamma }{2}}={\tfrac {\pi }{2}}.}
Combinando as dúas, conseguimos
A
2
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle A^{2}=s(s-a)(s-b)(s-c).}
A fórmula de Herón, como se indica anteriormente, é numericamente inestable para triángulos cun ángulo moi pequeno cando se usa a aritmética de coma flotante. Unha alternativa estable consiste en organizar as lonxitudes dos lados de xeito que
a
≥
b
≥
c
{\displaystyle a\geq b\geq c}
[ 5] [ 6]
A
=
1
4
(
a
+
(
b
+
c
)
)
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
a
+
(
b
−
c
)
)
.
{\displaystyle A={\tfrac {1}{4}}{\sqrt {{\big (}a+(b+c){\big )}{\big (}c-(a-b){\big )}{\big (}c+(a-b){\big )}{\big (}a+(b-c){\big )}}}.}
Respectando os corchetes na avaliación.
Fórmulas semellantes da área do triángulo [ editar | editar a fonte ] En función das medianas ,
m
a
,
{\displaystyle m_{a},}
m
b
,
{\displaystyle m_{b},}
m
c
{\displaystyle m_{c}}
σ
=
1
2
(
m
a
+
m
b
+
m
c
)
,
{\displaystyle \sigma ={\tfrac {1}{2}}(m_{a}+m_{b}+m_{c}),}
[ 7]
A
=
4
3
σ
(
σ
−
m
a
)
(
σ
−
m
b
)
(
σ
−
m
c
)
.
{\displaystyle A={\frac {4}{3}}{\sqrt {\sigma (\sigma -m_{a})(\sigma -m_{b})(\sigma -m_{c})}}.}
En función das alturas,
h
a
{\displaystyle h_{a}}
h
b
{\displaystyle h_{b}}
h
c
{\displaystyle h_{c}}
H
=
1
2
(
h
a
−
1
+
h
b
−
1
+
h
c
−
1
)
,
{\displaystyle H={\tfrac {1}{2}}{\bigl (}h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1}{\bigr )},}
[ 8]
A
−
1
=
4
H
(
H
−
h
a
−
1
)
(
H
−
h
b
−
1
)
(
H
−
h
c
−
1
)
.
{\displaystyle A^{-1}=4{\sqrt {H{\bigl (}H-h_{a}^{-1}{\bigr )}{\bigl (}H-h_{b}^{-1}{\bigr )}{\bigl (}H-h_{c}^{-1}{\bigr )}}}.}
En función dos ángulos
α
,
{\displaystyle \alpha ,}
β
,
{\displaystyle \beta ,}
γ
{\displaystyle \gamma }
S
=
1
2
(
sin
α
+
sin
β
+
sin
γ
)
,
{\displaystyle S={\tfrac {1}{2}}(\sin \alpha +\sin \beta +\sin \gamma ),}
[ 9] [ 10]
A
=
D
2
S
(
S
−
sin
α
)
(
S
−
sin
β
)
(
S
−
sin
γ
)
=
1
2
D
2
sin
α
sin
β
sin
γ
,
{\displaystyle {\begin{aligned}A&=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}\\[5mu]&={\tfrac {1}{2}}D^{2}\sin \alpha \,\sin \beta \,\sin \gamma ,\end{aligned}}}
onde
D
{\displaystyle D}
círculo circunscrito ,
D
=
a
/
sin
α
=
b
/
sin
β
=
c
/
sin
γ
.
{\displaystyle D=a/{\sin \alpha }=b/{\sin \beta }=c/{\sin \gamma }.}
Cuadrilátero cíclico A fórmula de Herón é un caso especial da fórmula de Brahmagupta para a área dun cuadrilátero cíclico . A fórmula de Herón e a fórmula de Brahmagupta son casos especiais da fórmula de Bretschneider para a área dun cuadrilátero . A fórmula de Herón pódese obter a partir da fórmula de Brahmagupta ou da fórmula de Bretschneider poñendo un dos lados do cuadrilátero a cero.
A fórmula de Brahmagupta para a área
K
{\displaystyle K}
a
,
{\displaystyle a,}
b
,
{\displaystyle b,}
c
,
{\displaystyle c,}
d
{\displaystyle d}
K
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
(
s
−
d
)
{\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}
onde
s
=
1
2
(
a
+
b
+
c
+
d
)
{\displaystyle s={\tfrac {1}{2}}(a+b+c+d)}
Outra xeneralización da fórmula de Herón para pentágonos e hexágonos inscritos nun círculo foi descuberta por David P. Robbins. [ 11]
Fórmula tipo Herón para o volume dun tetraedro [ editar | editar a fonte ] Se
U
,
{\displaystyle U,}
V
,
{\displaystyle V,}
W
,
{\displaystyle W,}
u
,
{\displaystyle u,}
v
,
{\displaystyle v,}
w
{\displaystyle w}
u
{\displaystyle u}
U
{\displaystyle U}
[ 12]
volume
=
(
−
a
+
b
+
c
+
d
)
(
a
−
b
+
c
+
d
)
(
a
+
b
−
c
+
d
)
(
a
+
b
+
c
−
d
)
192
u
v
w
{\displaystyle {\text{volume}}={\frac {\sqrt {\,(-a+b+c+d)\,(a-b+c+d)\,(a+b-c+d)\,(a+b+c-d)}}{192\,u\,v\,w}}}
onde
a
=
x
Y
Z
b
=
y
Z
X
c
=
z
X
Y
d
=
x
y
z
X
=
(
w
−
U
+
v
)
(
U
+
v
+
w
)
x
=
(
U
−
v
+
w
)
(
v
−
w
+
U
)
Y
=
(
u
−
V
+
w
)
(
V
+
w
+
u
)
y
=
(
V
−
w
+
u
)
(
w
−
u
+
V
)
Z
=
(
v
−
W
+
u
)
(
W
+
u
+
v
)
z
=
(
W
−
u
+
v
)
(
u
−
v
+
W
)
.
{\displaystyle {\begin{aligned}a&={\sqrt {xYZ}}\\b&={\sqrt {yZX}}\\c&={\sqrt {zXY}}\\d&={\sqrt {xyz}}\\X&=(w-U+v)\,(U+v+w)\\x&=(U-v+w)\,(v-w+U)\\Y&=(u-V+w)\,(V+w+u)\\y&=(V-w+u)\,(w-u+V)\\Z&=(v-W+u)\,(W+u+v)\\z&=(W-u+v)\,(u-v+W).\end{aligned}}}
Fórmulas de Herón en xeometrías non euclidianas [ editar | editar a fonte ] Tamén hai fórmulas para a área dun triángulo en función da lonxitude dos seus lados para os triángulos da esfera ou do plano hiperbólico . [ 13]
a
,
{\displaystyle a,}
b
,
{\displaystyle b,}
c
{\displaystyle c}
s
=
1
2
(
a
+
b
+
c
)
{\displaystyle s={\tfrac {1}{2}}(a+b+c)}
S
{\displaystyle S}
triángulo na esfera:
tan
2
S
4
=
tan
s
2
tan
s
−
a
2
tan
s
−
b
2
tan
s
−
c
2
{\displaystyle \tan ^{2}{\frac {S}{4}}=\tan {\frac {s}{2}}\tan {\frac {s-a}{2}}\tan {\frac {s-b}{2}}\tan {\frac {s-c}{2}}}
triángulo no plano hiperbólico:
tan
2
S
4
=
tanh
s
2
tanh
s
−
a
2
tanh
s
−
b
2
tanh
s
−
c
2
.
{\displaystyle \tan ^{2}{\frac {S}{4}}=\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}.}
↑ Kendig, Keith (2000). "Is a 2000-year-old formula still keeping some secrets?" . The American Mathematical Monthly 107 (5): 402–415. JSTOR 2695295 . MR 1763392 . doi :10.1080/00029890.2000.12005213 . ↑ Havel, Timothy F. (1991). "Some examples of the use of distances as coordinates for Euclidean geometry". Journal of Symbolic Computation 11 (5-6): 579–593. doi :10.1016/S0747-7171(08)80120-4 . ↑ Niven, Ivan (1981). Maxima and Minima Without Calculus 7–8 . ↑ Raifaizen, Claude H. (1971). A Simpler Proof of Heron's Formula . Mathematics Magazine 44 . pp. 27–28. doi :10.1080/0025570X.1971.11976093 . ↑ Sterbenz, Pat H. (1974-05-01). Floating-Point Computation . Prentice-Hall Series in Automatic Computation (1st ed.). Englewood Cliffs, New Jersey, USA: Prentice Hall . ISBN 0-13-322495-3 . ↑ William M. Kahan (24 March 2000). "Miscalculating Area and Angles of a Needle-like Triangle" (PDF) . ↑ Benyi, Arpad, "A Heron-type formula for the triangle," Mathematical Gazette 87, July 2003, 324–326.
↑ Mitchell, Douglas W., "A Heron-type formula for the reciprocal area of a triangle," Mathematical Gazette 89, November 2005, 494.
↑ Mitchell, Douglas W. (2009). A Heron-type area formula in terms of sines . Mathematical Gazette 93 . pp. 108–109. doi :10.1017/S002555720018430X . ↑ Kocik, Jerzy; Solecki, Andrzej (2009). Disentangling a triangle (PDF) . American Mathematical Monthly 116 . pp. 228–237. doi :10.1080/00029890.2009.11920932 . ↑ D. P. Robbins, "Areas of Polygons Inscribed in a Circle", Discr. Comput. Geom. 12, 223-236, 1994.
↑ W. Kahan, "What has the Volume of a Tetrahedron to do with Computer Programming Languages?",, pp. 16–17.
↑ Alekseevskij, D. V.; Vinberg, E. B.; Solodovnikov, A. S. (1993). "Geometry of spaces of constant curvature". En Gamkrelidze, R. V.; Vinberg, E. B. Geometry. II: Spaces of constant curvature . Encycl. Math. Sci. 29 . Springer-Verlag. p. 66. ISBN 1-56085-072-8 .
![{\displaystyle {\begin{aligned}A&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}.\\[6mu]A&={\tfrac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a71cf4290d1a89f3de7358dd71a23414b4855211)
![{\displaystyle {\begin{aligned}A&={\tfrac {1}{2}}({\mbox{base}})({\mbox{altura}})\\[6mu]&={\tfrac {1}{2}}ab\sin \gamma \\[6mu]&={\frac {ab}{4ab}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {-a^{4}-b^{4}-c^{4}+2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}}}\\[6mu]&={\tfrac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}\\[6mu]&={\sqrt {\left({\frac {a+b+c}{2}}\right)\left({\frac {-a+b+c}{2}}\right)\left({\frac {a-b+c}{2}}\right)\left({\frac {a+b-c}{2}}\right)}}\\[6mu]&={\sqrt {s(s-a)(s-b)(s-c)}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58aec2602f779f413cbba08c9183888c53ec27b4)
![{\displaystyle {\begin{aligned}A&=r(s-a)+r(s-b)+r(s-c)\\[2mu]&=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\[2mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left(\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\right)\\[3mu]&=r^{2}\left({\frac {s-a}{r}}\cdot {\frac {s-b}{r}}\cdot {\frac {s-c}{r}}\right)\\[3mu]&={\frac {(s-a)(s-b)(s-c)}{r}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c8fda08d86253fc84256b8229ec47df29418689)
![{\displaystyle {\begin{aligned}A&=D^{2}{\sqrt {S(S-\sin \alpha )(S-\sin \beta )(S-\sin \gamma )}}\\[5mu]&={\tfrac {1}{2}}D^{2}\sin \alpha \,\sin \beta \,\sin \gamma ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dd202793a39af6fd48df60bf7174e51e70a99710)
