tinka tik tiems trikampiams, pas kuriuos kampai prie pagrindo vienodi. Duotas trikampis, kurio pagrindas yra kraštinė c . Kairė kraštinė yra a . Dešinė kraštinė yra b . Be to,
b
>
a
.
{\displaystyle b>a.}
Į pagrindą c nuleista aukštinė h , kuri dalina pagrindą c į dvi atkarpas
c
a
{\displaystyle c_{a}}
c
b
{\displaystyle c_{b}}
c
b
>
c
a
.
{\displaystyle c_{b}>c_{a}.}
Akivaizdu, kad
b
a
=
c
b
c
a
.
{\displaystyle {\frac {b}{a}}={\frac {c_{b}}{c_{a}}}.}
Kadangi mes žinome visų trikampio kraštinių ilgius, tai pagal santykį tarp b ir a galime surasti ilgius
c
a
{\displaystyle c_{a}}
c
b
{\displaystyle c_{b}}
c ilgis.
Taigi randame,
c
b
=
b
a
b
a
+
1
⋅
c
,
{\displaystyle c_{b}={\frac {\frac {b}{a}}{{\frac {b}{a}}+1}}\cdot c,}
c
a
=
1
b
a
+
1
⋅
c
,
{\displaystyle c_{a}={\frac {1}{{\frac {b}{a}}+1}}\cdot c,}
kadangi
m
=
b
/
a
>
1
,
{\displaystyle m=b/a>1,}
m :1, nes
a
/
a
=
1.
{\displaystyle a/a=1.}
a
=
5
,
b
=
8
,
{\displaystyle a=5,\;b=8,}
c
b
:
c
a
=
1.6
:
1.
{\displaystyle c_{b}:c_{a}=1.6:1.}
c ilgį, tai
c
b
=
1.6
⋅
c
1.6
+
1
,
c
a
=
1
⋅
c
1.6
+
1
.
{\displaystyle c_{b}={\frac {1.6\cdot c}{1.6+1}},\;c_{a}={\frac {1\cdot c}{1.6+1}}.}
Žinodami kam lygi atkarpa
c
b
{\displaystyle c_{b}}
c
a
{\displaystyle c_{a}}
a , b , c , plotą. Nes tada galime rasti aukštinę h taikydami Pitagoro teoremą
h
=
b
2
−
c
b
2
{\displaystyle h={\sqrt {b^{2}-c_{b}^{2}}}}
h
=
a
2
−
c
a
2
.
{\displaystyle h={\sqrt {a^{2}-c_{a}^{2}}}.}
a , b , c yra
S
=
1
2
⋅
c
⋅
h
=
1
2
⋅
c
⋅
a
2
−
c
a
2
=
1
2
⋅
c
⋅
a
2
−
c
2
(
b
a
+
1
)
2
.
{\displaystyle S={\frac {1}{2}}\cdot c\cdot h={\frac {1}{2}}\cdot c\cdot {\sqrt {a^{2}-c_{a}^{2}}}={\frac {1}{2}}\cdot c\cdot {\sqrt {a^{2}-{\frac {c^{2}}{({\frac {b}{a}}+1)^{2}}}}}.}
Herono formulės įrodymas
Turime trikampį ABC ir turime tokias kraštines CB=a, AB=c, AC=b. Iš kampo A nuleidžiame aukštinę h į ilgiausią trikampio kraštinę a , o toje vietoje kur susikerta aukštinė h su kraštine a yra taškas D . Tuomet DB=x ir CD=a-x. Iš pitagoro teoremos žinome, kad
h
=
c
2
−
x
2
,
{\displaystyle h={\sqrt {c^{2}-x^{2}}},}
(
a
−
x
)
2
=
b
2
−
h
2
.
{\displaystyle (a-x)^{2}=b^{2}-h^{2}.}
Tuomet
h
{\displaystyle h}
(
a
−
x
)
2
=
b
2
−
(
c
2
−
x
2
)
2
,
{\displaystyle (a-x)^{2}=b^{2}-({\sqrt {c^{2}-x^{2}}})^{2},}
a
2
−
2
a
x
+
x
2
=
b
2
−
(
c
2
−
x
2
)
,
{\displaystyle a^{2}-2ax+x^{2}=b^{2}-(c^{2}-x^{2}),}
a
2
−
2
a
x
+
x
2
=
b
2
−
c
2
+
x
2
,
{\displaystyle a^{2}-2ax+x^{2}=b^{2}-c^{2}+x^{2},}
a
2
−
2
a
x
=
b
2
−
c
2
,
{\displaystyle a^{2}-2ax=b^{2}-c^{2},}
−
2
a
x
=
b
2
−
c
2
−
a
2
,
{\displaystyle -2ax=b^{2}-c^{2}-a^{2},}
2
a
x
=
c
2
+
a
2
−
b
2
,
{\displaystyle 2ax=c^{2}+a^{2}-b^{2},}
x
=
c
2
+
a
2
−
b
2
2
a
.
{\displaystyle x={\frac {c^{2}+a^{2}-b^{2}}{2a}}.}
Randame Trikampio ABC aukšinę:
h
=
c
2
−
x
2
=
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
=
c
2
−
(
c
2
+
a
2
−
b
2
)
2
4
a
2
=
c
2
−
(
c
4
+
a
2
c
2
−
b
2
c
2
)
+
(
a
2
c
2
+
a
4
−
a
2
b
2
)
−
b
2
c
2
−
a
2
b
2
+
b
4
4
a
2
=
{\displaystyle h={\sqrt {c^{2}-x^{2}}}={\sqrt {c^{2}-\left({\frac {c^{2}+a^{2}-b^{2}}{2a}}\right)^{2}}}={\sqrt {c^{2}-{\frac {(c^{2}+a^{2}-b^{2})^{2}}{4a^{2}}}}}={\sqrt {c^{2}-{\frac {(c^{4}+a^{2}c^{2}-b^{2}c^{2})+(a^{2}c^{2}+a^{4}-a^{2}b^{2})-b^{2}c^{2}-a^{2}b^{2}+b^{4}}{4a^{2}}}}}=}
=
c
2
−
c
4
+
2
a
2
c
2
−
2
b
2
c
2
+
a
4
−
2
a
2
b
2
+
b
4
4
a
2
=
4
a
2
c
2
−
(
c
4
+
2
a
2
c
2
−
2
b
2
c
2
+
a
4
−
2
a
2
b
2
+
b
4
)
4
a
2
=
{\displaystyle ={\sqrt {c^{2}-{\frac {c^{4}+2a^{2}c^{2}-2b^{2}c^{2}+a^{4}-2a^{2}b^{2}+b^{4}}{4a^{2}}}}}={\sqrt {\frac {4a^{2}c^{2}-(c^{4}+2a^{2}c^{2}-2b^{2}c^{2}+a^{4}-2a^{2}b^{2}+b^{4})}{4a^{2}}}}=}
=
4
a
2
c
2
−
c
4
−
2
a
2
c
2
+
2
b
2
c
2
−
a
4
+
2
a
2
b
2
−
b
4
4
a
2
=
2
a
2
c
2
−
c
4
+
2
b
2
c
2
−
a
4
+
2
a
2
b
2
−
b
4
4
a
2
=
2
(
a
2
c
2
+
b
2
c
2
+
a
2
b
2
)
−
c
4
−
a
4
−
b
4
4
a
2
=
{\displaystyle ={\sqrt {\frac {4a^{2}c^{2}-c^{4}-2a^{2}c^{2}+2b^{2}c^{2}-a^{4}+2a^{2}b^{2}-b^{4}}{4a^{2}}}}={\sqrt {\frac {2a^{2}c^{2}-c^{4}+2b^{2}c^{2}-a^{4}+2a^{2}b^{2}-b^{4}}{4a^{2}}}}={\sqrt {\frac {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}{4a^{2}}}}=}
=
−
(
−
2
a
2
c
2
−
2
b
2
c
2
−
2
a
2
b
2
+
c
4
+
a
4
+
b
4
)
4
a
2
=
−
(
c
2
(
−
2
a
2
−
2
b
2
+
c
2
)
+
(
a
2
−
b
2
)
2
)
4
a
2
=
c
2
(
2
(
a
2
+
b
2
)
−
c
2
)
−
(
a
2
−
b
2
)
2
2
a
.
{\displaystyle ={\sqrt {\frac {-(-2a^{2}c^{2}-2b^{2}c^{2}-2a^{2}b^{2}+c^{4}+a^{4}+b^{4})}{4a^{2}}}}={\sqrt {\frac {-(c^{2}(-2a^{2}-2b^{2}+c^{2})+(a^{2}-b^{2})^{2})}{4a^{2}}}}={\frac {\sqrt {c^{2}(2(a^{2}+b^{2})-c^{2})-(a^{2}-b^{2})^{2}}}{2a}}.}
h
=
c
2
−
x
2
=
(
c
−
x
)
(
c
+
x
)
=
(
c
−
c
2
+
a
2
−
b
2
2
a
)
(
c
+
c
2
+
a
2
−
b
2
2
a
)
=
(
2
a
c
−
(
c
2
+
a
2
−
b
2
)
2
a
)
(
2
a
c
+
c
2
+
a
2
−
b
2
2
a
)
=
{\displaystyle h={\sqrt {c^{2}-x^{2}}}={\sqrt {(c-x)(c+x)}}={\sqrt {(c-{\frac {c^{2}+a^{2}-b^{2}}{2a}})(c+{\frac {c^{2}+a^{2}-b^{2}}{2a}})}}={\sqrt {({\frac {2ac-(c^{2}+a^{2}-b^{2})}{2a}})({\frac {2ac+c^{2}+a^{2}-b^{2}}{2a}})}}=}
=
(
2
a
c
−
c
2
−
a
2
+
b
2
2
a
)
(
(
c
+
a
)
2
−
b
2
2
a
)
=
(
−
(
a
2
−
2
a
c
+
c
2
)
+
b
2
2
a
)
(
(
(
c
+
a
)
−
b
)
(
(
c
+
a
)
+
b
)
2
a
)
=
{\displaystyle ={\sqrt {({\frac {2ac-c^{2}-a^{2}+b^{2}}{2a}})({\frac {(c+a)^{2}-b^{2}}{2a}})}}={\sqrt {({\frac {-(a^{2}-2ac+c^{2})+b^{2}}{2a}})({\frac {((c+a)-b)((c+a)+b)}{2a}})}}=}
=
(
b
2
−
(
a
−
c
)
2
2
a
)
(
(
c
+
a
−
b
)
(
c
+
a
+
b
)
2
a
)
=
(
(
b
−
(
a
−
c
)
)
(
b
+
(
a
−
c
)
)
2
a
)
(
(
c
+
a
−
b
)
(
c
+
a
+
b
)
2
a
)
=
{\displaystyle ={\sqrt {({\frac {b^{2}-(a-c)^{2}}{2a}})({\frac {(c+a-b)(c+a+b)}{2a}})}}={\sqrt {({\frac {(b-(a-c))(b+(a-c))}{2a}})({\frac {(c+a-b)(c+a+b)}{2a}})}}=}
=
(
(
b
−
a
+
c
)
(
b
+
a
−
c
)
2
a
)
(
(
c
+
a
−
b
)
(
c
+
a
+
b
)
2
a
)
=
(
b
−
a
+
c
)
(
b
+
a
−
c
)
(
c
+
a
−
b
)
2
p
4
a
2
=
2
(
p
−
a
)
2
(
p
−
c
)
2
(
p
−
b
)
2
p
4
a
2
=
{\displaystyle ={\sqrt {({\frac {(b-a+c)(b+a-c)}{2a}})({\frac {(c+a-b)(c+a+b)}{2a}})}}={\sqrt {\frac {(b-a+c)(b+a-c)(c+a-b)2p}{4a^{2}}}}={\sqrt {\frac {2(p-a)2(p-c)2(p-b)2p}{4a^{2}}}}=}
=
16
p
(
p
−
a
)
(
p
−
c
)
(
p
−
b
)
4
a
2
=
4
p
(
p
−
a
)
(
p
−
c
)
(
p
−
b
)
a
2
.
{\displaystyle ={\sqrt {\frac {16p(p-a)(p-c)(p-b)}{4a^{2}}}}={\sqrt {\frac {4p(p-a)(p-c)(p-b)}{a^{2}}}}.}
p=(a+b+c)/2, 2p=(a+b+c), 2p-2a=a+b+c-2a=-a+b+c, 2(p-a)=-a+b+c ir taip pat su kitais. Dabar galime surasti trikampio plotą:
S
Δ
A
B
C
=
1
2
⋅
a
⋅
h
=
1
2
⋅
a
⋅
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
.
{\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot a\cdot h={\frac {1}{2}}\cdot a\cdot {\sqrt {c^{2}-\left({\frac {c^{2}+a^{2}-b^{2}}{2a}}\right)^{2}}}.}
S
Δ
A
B
C
=
1
2
⋅
a
⋅
h
=
1
2
⋅
a
⋅
2
(
a
2
c
2
+
b
2
c
2
+
a
2
b
2
)
−
c
4
−
a
4
−
b
4
4
a
2
=
1
4
⋅
2
(
a
2
c
2
+
b
2
c
2
+
a
2
b
2
)
−
c
4
−
a
4
−
b
4
.
{\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot a\cdot h={\frac {1}{2}}\cdot a\cdot {\sqrt {\frac {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}{4a^{2}}}}={\frac {1}{4}}\cdot {\sqrt {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}.}
Pavyzdis, kai a=6, h=4, b=5, c=5, tai S=a*h/2=6*4/2=12. O taip pat ir:
S
Δ
=
1
2
⋅
a
⋅
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
=
1
2
⋅
6
⋅
5
2
−
(
5
2
+
6
2
−
5
2
2
⋅
6
)
2
=
3
⋅
25
−
(
25
+
36
−
25
12
)
2
=
3
⋅
25
−
(
36
12
)
2
=
{\displaystyle S_{\Delta }={\frac {1}{2}}\cdot a\cdot {\sqrt {c^{2}-\left({\frac {c^{2}+a^{2}-b^{2}}{2a}}\right)^{2}}}={\frac {1}{2}}\cdot 6\cdot {\sqrt {5^{2}-\left({\frac {5^{2}+6^{2}-5^{2}}{2\cdot 6}}\right)^{2}}}=3\cdot {\sqrt {25-\left({\frac {25+36-25}{12}}\right)^{2}}}=3\cdot {\sqrt {25-\left({\frac {36}{12}}\right)^{2}}}=}
=
3
⋅
25
−
3
2
=
3
⋅
25
−
9
=
3
⋅
16
=
3
⋅
4
=
12.
{\displaystyle =3\cdot {\sqrt {25-3^{2}}}=3\cdot {\sqrt {25-9}}=3\cdot {\sqrt {16}}=3\cdot 4=12.}
Pavyzdis . Duotas status trikampis ABC, kurio vienas statinis yra a=3, o kitas statinis yra b=h=4, o įžambinė c=5. Trikampio plotas yra S=a*b/2=3*4/2=12/2=6. Dabar rasime ši plotą per Herono formulę:
S
Δ
A
B
C
=
1
2
⋅
a
⋅
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
=
1
2
⋅
3
⋅
5
2
−
(
5
2
+
3
2
−
4
2
2
⋅
3
)
2
=
3
2
⋅
25
−
(
25
+
9
−
16
6
)
2
=
3
2
⋅
25
−
(
18
6
)
2
=
{\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot a\cdot {\sqrt {c^{2}-\left({\frac {c^{2}+a^{2}-b^{2}}{2a}}\right)^{2}}}={\frac {1}{2}}\cdot 3\cdot {\sqrt {5^{2}-\left({\frac {5^{2}+3^{2}-4^{2}}{2\cdot 3}}\right)^{2}}}={\frac {3}{2}}\cdot {\sqrt {25-\left({\frac {25+9-16}{6}}\right)^{2}}}={\frac {3}{2}}\cdot {\sqrt {25-\left({\frac {18}{6}}\right)^{2}}}=}
=
3
2
⋅
25
−
3
2
=
3
2
⋅
25
−
9
=
3
2
⋅
16
=
3
2
⋅
4
=
3
⋅
2
=
6.
{\displaystyle ={\frac {3}{2}}\cdot {\sqrt {25-3^{2}}}={\frac {3}{2}}\cdot {\sqrt {25-9}}={\frac {3}{2}}\cdot {\sqrt {16}}={\frac {3}{2}}\cdot 4=3\cdot 2=6.}
S
Δ
A
B
C
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
6
(
6
−
3
)
(
6
−
4
)
(
6
−
3
)
=
6
⋅
3
⋅
2
⋅
1
=
36
=
6
,
{\displaystyle S_{\Delta ABC}={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {6(6-3)(6-4)(6-3)}}={\sqrt {6\cdot 3\cdot 2\cdot 1}}={\sqrt {36}}=6,}
p
=
P
2
=
a
+
b
+
c
2
=
3
+
4
+
5
2
=
12
2
=
6.
{\displaystyle p={\frac {P}{2}}={\frac {a+b+c}{2}}={\frac {3+4+5}{2}}={\frac {12}{2}}=6.}
![{\displaystyle ={\frac {1}{4}}{\sqrt {2([(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}][(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2}]+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52855110e143a93169c6353b91b52bcf3edf7022)
