Išvesime, trikampio erdvėje ploto radimo formulę. Trikampį ABC sudaro taškai
A
(
x
1
;
y
1
;
z
1
)
{\displaystyle A(x_{1};y_{1};z_{1})}
B
(
x
2
;
y
2
;
z
2
)
{\displaystyle B(x_{2};y_{2};z_{2})}
C
(
x
3
;
y
3
;
z
3
)
{\displaystyle C(x_{3};y_{3};z_{3})}
ABC Kraštinių ilgiai yra:
a
=
A
B
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
+
(
z
2
−
z
1
)
2
;
{\displaystyle a=AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}};}
b
=
A
C
=
(
x
3
−
x
1
)
2
+
(
y
3
−
y
1
)
2
+
(
z
3
−
z
1
)
2
;
{\displaystyle b=AC={\sqrt {(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}+(z_{3}-z_{1})^{2}}};}
c
=
B
C
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
+
(
z
3
−
z
2
)
2
.
{\displaystyle c=BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2}}}.}
a
2
=
A
B
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
+
(
z
2
−
z
1
)
2
=
x
2
2
−
2
x
2
x
1
+
x
1
2
+
y
2
2
−
2
y
2
y
1
+
y
1
2
+
z
2
2
−
2
z
2
z
1
+
z
1
2
;
{\displaystyle a^{2}=AB^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}=x_{2}^{2}-2x_{2}x_{1}+x_{1}^{2}+y_{2}^{2}-2y_{2}y_{1}+y_{1}^{2}+z_{2}^{2}-2z_{2}z_{1}+z_{1}^{2};}
b
2
=
A
C
2
=
(
x
3
−
x
1
)
2
+
(
y
3
−
y
1
)
2
+
(
z
3
−
z
1
)
2
=
x
3
2
−
2
x
3
x
1
+
x
1
2
+
y
3
2
−
2
y
3
y
1
+
y
1
2
+
z
3
2
−
2
z
3
z
1
+
z
1
2
;
{\displaystyle b^{2}=AC^{2}=(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}+(z_{3}-z_{1})^{2}=x_{3}^{2}-2x_{3}x_{1}+x_{1}^{2}+y_{3}^{2}-2y_{3}y_{1}+y_{1}^{2}+z_{3}^{2}-2z_{3}z_{1}+z_{1}^{2};}
c
2
=
B
C
2
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
+
(
z
3
−
z
2
)
2
=
x
3
2
−
2
x
3
x
2
+
x
2
2
+
y
3
2
−
2
y
3
y
2
+
y
2
2
+
z
3
2
−
2
z
3
z
2
+
z
2
2
.
{\displaystyle c^{2}=BC^{2}=(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2}=x_{3}^{2}-2x_{3}x_{2}+x_{2}^{2}+y_{3}^{2}-2y_{3}y_{2}+y_{2}^{2}+z_{3}^{2}-2z_{3}z_{2}+z_{2}^{2}.}
Iš Herono formulės turime išvedę, kad trikampio ABC plotas yra:
S
Δ
A
B
C
=
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}{4}}\cdot {\sqrt {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}.}
S
Δ
A
B
C
=
1
2
⋅
a
⋅
h
=
1
2
⋅
a
⋅
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
=
1
4
⋅
4
a
2
c
2
−
(
c
2
+
a
2
−
b
2
)
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}}}={\frac {1}{4}}\cdot {\sqrt {4a^{2}c^{2}-(c^{2}+a^{2}-b^{2})^{2}}}.}
Taigi turime:
S
Δ
A
B
C
=
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}{4}}{\sqrt {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}=}
=
1
4
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
=
{\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}}}=}
PIRMAS METODAS: Duotas trikampis, kurio pagrindas
c
=
30.
{\displaystyle c=30.}
a
=
20.
{\displaystyle a=20.}
b
=
25.
{\displaystyle b=25.}
Rasti trikampio sudaryto iš kraštinių a, b, c plotą.
Sprendimas .
p
=
a
+
b
+
c
2
=
30
+
25
+
20
2
=
75
2
=
37.5
;
{\displaystyle p={\frac {a+b+c}{2}}={\frac {30+25+20}{2}}={\frac {75}{2}}=37.5;}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
37.5
(
37.5
−
30
)
(
37.5
−
25
)
(
37.5
−
20
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {37.5(37.5-30)(37.5-25)(37.5-20)}}=}
=
37.5
⋅
7.5
⋅
12.5
⋅
17.5
=
37.5
⋅
7.5
⋅
12.5
⋅
17.5
=
{\displaystyle ={\sqrt {37.5\cdot 7.5\cdot 12.5\cdot 17.5}}={\sqrt {37.5\cdot 7.5\cdot 12.5\cdot 17.5}}=}
=
61523.4375
=
248.0391854.
{\displaystyle ={\sqrt {61523.4375}}=248.0391854.}
ANTRAS METODAS: Duotas trikampis, kurio pagrindas
c
=
30.
{\displaystyle c=30.}
a
=
20.
{\displaystyle a=20.}
b
=
25.
{\displaystyle b=25.}
Rasti trikampio sudaryto iš kraštinių a, b, c plotą.
Sprendimas .
S
=
1
2
⋅
c
⋅
a
2
−
c
2
(
b
a
+
1
)
2
=
1
2
⋅
30
⋅
20
2
−
30
2
(
25
20
+
1
)
2
=
{\displaystyle S={\frac {1}{2}}\cdot c\cdot {\sqrt {a^{2}-{\frac {c^{2}}{({\frac {b}{a}}+1)^{2}}}}}={\frac {1}{2}}\cdot 30\cdot {\sqrt {20^{2}-{\frac {30^{2}}{({\frac {25}{20}}+1)^{2}}}}}=}
=
15
400
−
900
(
5
4
+
1
)
2
=
15
400
−
900
(
5
+
4
4
)
2
=
15
400
−
900
(
9
4
)
2
=
{\displaystyle =15{\sqrt {400-{\frac {900}{({\frac {5}{4}}+1)^{2}}}}}=15{\sqrt {400-{\frac {900}{({\frac {5+4}{4}})^{2}}}}}=15{\sqrt {400-{\frac {900}{({\frac {9}{4}})^{2}}}}}=}
=
15
400
−
900
81
16
=
15
400
−
100
⋅
16
9
=
15
400
⋅
9
−
1600
9
=
{\displaystyle =15{\sqrt {400-{\frac {900}{\frac {81}{16}}}}}=15{\sqrt {400-{\frac {100\cdot 16}{9}}}}=15{\sqrt {\frac {400\cdot 9-1600}{9}}}=}
=
15
3
3600
−
1600
=
5
2000
=
223.6067977.
{\displaystyle ={\frac {15}{3}}{\sqrt {3600-1600}}=5{\sqrt {2000}}=223.6067977.}
Pirmas metodas teisingas, o antras metodas neteisingas. Toks trikampis gali būti. Negali būti tik toks trikampis, kurio ilgiausia kraštinė ilgesnė (arba lygi) už dvejų kitų trikampio kraštinių ilgių sumą. 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}}}}}.}
Duotas trikampis ABC , kurio visos kraštinės lygios
a
=
b
=
c
=
5.
{\displaystyle a=b=c=5.}
ABC plotą. Sprendimas .
S
=
1
2
⋅
c
⋅
a
2
−
c
2
(
b
a
+
1
)
2
=
1
2
⋅
5
⋅
5
2
−
5
2
(
5
5
+
1
)
2
=
{\displaystyle S={\frac {1}{2}}\cdot c\cdot {\sqrt {a^{2}-{\frac {c^{2}}{({\frac {b}{a}}+1)^{2}}}}}={\frac {1}{2}}\cdot 5\cdot {\sqrt {5^{2}-{\frac {5^{2}}{({\frac {5}{5}}+1)^{2}}}}}=}
=
1
2
⋅
5
⋅
25
−
25
4
=
1
2
⋅
5
⋅
100
−
25
4
=
{\displaystyle ={\frac {1}{2}}\cdot 5\cdot {\sqrt {25-{\frac {25}{4}}}}={\frac {1}{2}}\cdot 5\cdot {\sqrt {\frac {100-25}{4}}}=}
=
5
2
75
4
=
25
3
4
=
10.82531755.
{\displaystyle ={\frac {5}{2}}{\sqrt {\frac {75}{4}}}={\frac {25{\sqrt {3}}}{4}}=10.82531755.}
Duotas trikampis, kurio pagrindas
c
=
30.
{\displaystyle c=30.}
a
=
20.
{\displaystyle a=20.}
b
=
25.
{\displaystyle b=25.}
Rasti trikampio sudaryto iš kraštinių a, b, c plotą.
Sprendimas .
S
=
1
2
⋅
c
⋅
a
2
−
c
2
(
b
a
+
1
)
2
=
1
2
⋅
30
⋅
20
2
−
30
2
(
25
20
+
1
)
2
=
{\displaystyle S={\frac {1}{2}}\cdot c\cdot {\sqrt {a^{2}-{\frac {c^{2}}{({\frac {b}{a}}+1)^{2}}}}}={\frac {1}{2}}\cdot 30\cdot {\sqrt {20^{2}-{\frac {30^{2}}{({\frac {25}{20}}+1)^{2}}}}}=}
=
15
400
−
900
(
5
4
+
1
)
2
=
15
400
−
900
(
5
+
4
4
)
2
=
15
400
−
900
(
9
4
)
2
=
{\displaystyle =15{\sqrt {400-{\frac {900}{({\frac {5}{4}}+1)^{2}}}}}=15{\sqrt {400-{\frac {900}{({\frac {5+4}{4}})^{2}}}}}=15{\sqrt {400-{\frac {900}{({\frac {9}{4}})^{2}}}}}=}
=
15
400
−
900
81
16
=
15
400
−
100
⋅
16
9
=
15
400
⋅
9
−
1600
9
=
{\displaystyle =15{\sqrt {400-{\frac {900}{\frac {81}{16}}}}}=15{\sqrt {400-{\frac {100\cdot 16}{9}}}}=15{\sqrt {\frac {400\cdot 9-1600}{9}}}=}
=
15
3
3600
−
1600
=
5
2000
=
223.6067977.
{\displaystyle ={\frac {15}{3}}{\sqrt {3600-1600}}=5{\sqrt {2000}}=223.6067977.}
a=AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}; b=AC=\sqrt{(x_3-x_1)^2+(y_3-y_1)^2+(z_3-z_1)^2}; c=BC=\sqrt{(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2}. a^2=AB^2=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2=x_2^2-2 x_2 x_1+x_1^2+y_2^2-2 y_2 y_1+y_1^2+z_2^2-2 z_2 z_1+z_1^2; b^2=AC^2=(x_3-x_1)^2+(y_3-y_1)^2+(z_3-z_1)^2=x_3^2-2 x_3 x_1+x_1^2+y_3^2-2 y_3 y_1+y_1^2+z_3^2-2 z_3 z_1+z_1^2; c^2=BC^2=(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2=x_3^2-2 x_3 x_2+x_2^2+y_3^2-2 y_3 y_2+y_2^2+z_3^2-2 z_3 z_2+z_2^2. Iš Herono formulės turime išvedę, kad trikampio ABC plotas yra: S_{\Delta ABC}=\frac{1}{4}\cdot \sqrt{2 (a^2 c^2+ b^2 c^2+ a^2 b^2)-c^4-a^4-b^4}. 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}=\frac{1}{4}\cdot \sqrt{ 4a^2 c^2-(c^2+a^2-b^2)^2}. Taigi turime: S_{\Delta ABC}=\frac{1}{4}\sqrt{2 (a^2 c^2+ b^2 c^2+ a^2 b^2)-c^4-a^4-b^4}= =\frac{1}{4}\sqrt{2 ([(x_2+ b^2 c^2+ a^2 b^2)-c^4-a^4-b^4}= 18
a
=
A
B
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
+
(
z
2
−
z
1
)
2
;
{\displaystyle a=AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}};}
b
=
A
C
=
(
x
3
−
x
1
)
2
+
(
y
3
−
y
1
)
2
+
(
z
3
−
z
1
)
2
;
{\displaystyle b=AC={\sqrt {(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}+(z_{3}-z_{1})^{2}}};}
c
=
B
C
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
+
(
z
3
−
z
2
)
2
.
{\displaystyle c=BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2}}}.}
a
2
=
A
B
2
=
(
x
2
−
x
1
)
2
+
(
y
2
−
y
1
)
2
+
(
z
2
−
z
1
)
2
=
x
2
2
−
2
x
2
x
1
+
x
1
2
+
y
2
2
−
2
y
2
y
1
+
y
1
2
+
z
2
2
−
2
z
2
z
1
+
z
1
2
;
{\displaystyle a^{2}=AB^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}=x_{2}^{2}-2x_{2}x_{1}+x_{1}^{2}+y_{2}^{2}-2y_{2}y_{1}+y_{1}^{2}+z_{2}^{2}-2z_{2}z_{1}+z_{1}^{2};}
b
2
=
A
C
2
=
(
x
3
−
x
1
)
2
+
(
y
3
−
y
1
)
2
+
(
z
3
−
z
1
)
2
=
x
3
2
−
2
x
3
x
1
+
x
1
2
+
y
3
2
−
2
y
3
y
1
+
y
1
2
+
z
3
2
−
2
z
3
z
1
+
z
1
2
;
{\displaystyle b^{2}=AC^{2}=(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}+(z_{3}-z_{1})^{2}=x_{3}^{2}-2x_{3}x_{1}+x_{1}^{2}+y_{3}^{2}-2y_{3}y_{1}+y_{1}^{2}+z_{3}^{2}-2z_{3}z_{1}+z_{1}^{2};}
c
2
=
B
C
2
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
+
(
z
3
−
z
2
)
2
=
x
3
2
−
2
x
3
x
2
+
x
2
2
+
y
3
2
−
2
y
3
y
2
+
y
2
2
+
z
3
2
−
2
z
3
z
2
+
z
2
2
.
{\displaystyle c^{2}=BC^{2}=(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2}=x_{3}^{2}-2x_{3}x_{2}+x_{2}^{2}+y_{3}^{2}-2y_{3}y_{2}+y_{2}^{2}+z_{3}^{2}-2z_{3}z_{2}+z_{2}^{2}.}
Iš Herono formulės turime išvedę, kad trikampio ABC plotas yra:
S
Δ
A
B
C
=
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}{4}}\cdot {\sqrt {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}.}
S
Δ
A
B
C
=
1
2
⋅
a
⋅
h
=
1
2
⋅
a
⋅
c
2
−
(
c
2
+
a
2
−
b
2
2
a
)
2
=
1
4
⋅
4
a
2
c
2
−
(
c
2
+
a
2
−
b
2
)
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}}}={\frac {1}{4}}\cdot {\sqrt {4a^{2}c^{2}-(c^{2}+a^{2}-b^{2})^{2}}}.}
Taigi turime:
S
Δ
A
B
C
=
1
4
2
(
a
2
c
2
+
b
2
c
2
+
a
2
b
2
)
−
c
4
−
a
4
−
b
4
==
1
4
2
(
[
(
x
2
+
b
2
c
2
+
a
2
b
2
)
−
c
4
−
a
4
−
b
4
=
18
{\displaystyle S_{\Delta ABC}={\frac {1}{4}}{\sqrt {2(a^{2}c^{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}=={\frac {1}{4}}{\sqrt {2([(x_{2}+b^{2}c^{2}+a^{2}b^{2})-c^{4}-a^{4}-b^{4}}}=18}
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)
