Bretschneider's-ova formula se koristi u geometriji za određivanje površine četvorougla , i glasi
Četverougao
P
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
(
s
−
d
)
−
a
b
c
d
cos
2
α
+
γ
2
,
{\displaystyle P={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}}},}
pri čemu su, a , b , c i d stranice četvorougla, s je polovina obima četvorougla, a
α
{\displaystyle \alpha \,}
γ
{\displaystyle \gamma \,}
uglovi .
Bretschneider's-ova formula daje površinu četvorougla bez obzira da li je on tetivan ili nije.
Ako se površina četvorougla označi sa P , onda važi
P
=
P
△
B
D
C
+
P
△
A
D
B
=
1
2
a
b
sin
γ
+
1
2
c
d
sin
α
{\displaystyle {\begin{aligned}P=P_{\triangle BDC}+P_{\triangle ADB}={\tfrac {1}{2}}ab\sin \gamma +{\tfrac {1}{2}}cd\sin \alpha \end{aligned}}}
4
P
2
=
(
a
b
)
2
sin
2
γ
+
(
c
d
)
2
sin
2
α
+
2
a
b
c
d
sin
α
sin
γ
.
{\displaystyle 4P^{2}=(ab)^{2}\sin ^{2}\gamma +(cd)^{2}\sin ^{2}\alpha +2abcd\sin \alpha \sin \gamma .\,}
Prema kosinusnoj teoremi
a
2
+
b
2
−
2
a
b
cos
γ
=
c
2
+
d
2
−
2
c
d
cos
α
,
{\displaystyle a^{2}+b^{2}-2ab\cos \gamma =c^{2}+d^{2}-2cd\cos \alpha ,\,}
Pošto su obe strane jednake kvadratu dužine dijagonale BD i
1
4
(
c
2
+
d
2
−
a
2
−
b
2
)
2
=
(
a
b
)
2
cos
2
γ
+
(
c
d
)
2
cos
2
α
−
2
a
b
c
d
cos
α
cos
γ
.
{\displaystyle {\tfrac {1}{4}}(c^{2}+d^{2}-a^{2}-b^{2})^{2}=(ab)^{2}\cos ^{2}\gamma +(cd)^{2}\cos ^{2}\alpha -2abcd\cos \alpha \cos \gamma .\,}
4
P
2
+
1
4
(
c
2
+
d
2
−
a
2
−
b
2
)
2
=
(
a
b
)
2
+
(
c
d
)
2
−
2
a
b
c
d
cos
(
α
+
γ
)
.
{\displaystyle 4P^{2}+{\tfrac {1}{4}}(c^{2}+d^{2}-a^{2}-b^{2})^{2}=(ab)^{2}+(cd)^{2}-2abcd\cos(\alpha +\gamma ).\,}
16
P
2
=
4
(
a
2
b
2
+
c
2
d
2
)
−
(
c
2
+
d
2
−
a
2
−
b
2
)
2
−
8
a
b
c
d
cos
(
α
+
γ
)
.
{\displaystyle 16P^{2}=4(a^{2}b^{2}+c^{2}d^{2})-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd\cos(\alpha +\gamma ).\,}
16
P
2
=
4
(
a
b
+
c
d
)
2
−
(
c
2
+
d
2
−
a
2
−
b
2
)
2
−
8
a
b
c
d
[
1
+
cos
(
α
+
γ
)
]
.
{\displaystyle 16P^{2}=4(ab+cd)^{2}-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd[1+\cos(\alpha +\gamma )].\,}
16
P
2
=
[
2
(
a
b
+
c
d
)
−
(
c
2
+
d
2
−
a
2
−
b
2
)
]
[
2
(
a
b
+
c
d
)
+
(
c
2
+
d
2
−
a
2
−
b
2
)
]
−
16
a
b
c
d
cos
2
α
+
γ
2
,
{\displaystyle 16P^{2}=[2(ab+cd)-(c^{2}+d^{2}-a^{2}-b^{2})][2(ab+cd)+(c^{2}+d^{2}-a^{2}-b^{2})]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}},\,}
16
P
2
=
[
(
a
+
b
)
2
−
(
c
−
d
)
2
]
[
(
c
+
d
)
2
−
(
a
−
b
)
2
]
−
16
a
b
c
d
cos
2
α
+
γ
2
.
{\displaystyle 16P^{2}=[(a+b)^{2}-(c-d)^{2}][(c+d)^{2}-(a-b)^{2}]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.\,}
16
P
2
=
(
a
+
b
+
c
−
d
)
(
a
+
b
−
c
+
d
)
(
c
+
d
+
a
−
b
)
(
c
+
d
−
a
+
b
)
−
16
a
b
c
d
cos
2
α
+
γ
2
.
{\displaystyle 16P^{2}=(a+b+c-d)(a+b-c+d)(c+d+a-b)(c+d-a+b)-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.}
Uzevši u obzir za poluobim četverougla
s
=
a
+
b
+
c
+
d
2
,
{\displaystyle s={\frac {a+b+c+d}{2}},}
imamo
16
P
2
=
16
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
(
s
−
d
)
−
16
a
b
c
d
cos
2
α
+
γ
2
{\displaystyle 16P^{2}=16(s-a)(s-b)(s-c)(s-d)-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}}
Bretschneider's-ova formulaje uopštenje formule Bramagupte za površinu tetivnog četvorougla, a ova je uopštenje Heronovog obrasca koji se koristi za izračunavanje površine trougla.
![{\displaystyle 16P^{2}=4(ab+cd)^{2}-(c^{2}+d^{2}-a^{2}-b^{2})^{2}-8abcd[1+\cos(\alpha +\gamma )].\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c8522a330ad9e3791d58d10ff32485453eaf9d9)
![{\displaystyle 16P^{2}=[2(ab+cd)-(c^{2}+d^{2}-a^{2}-b^{2})][2(ab+cd)+(c^{2}+d^{2}-a^{2}-b^{2})]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}},\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6c6f1d7bd146e9bee1c1e5b0df9a7ed7e67a503d)
![{\displaystyle 16P^{2}=[(a+b)^{2}-(c-d)^{2}][(c+d)^{2}-(a-b)^{2}]-16abcd\cos ^{2}{\frac {\alpha +\gamma }{2}}.\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/049dd807f0ac5ef9be248dc9466c440bcbd2c004)
