ບົດຄວາມນີ້ກ່ຽວກັບcalculating the area of a triangle ສຳລັບcalculating a square root ເບິ່ງທີ່
Heron's method ໃນເລຂາຄະນິດ geometry, ສູດເຮລົງ Heron's (ຫຼື Hero's) formula , ແມ່ນຖືກຕັ້ງຊື່ຕາມ ເຮລົງ ອະເລັກຊານເດຍ (Heron of Alexandria), ໂດຍວ່າ ເນື້ອທີ່ S ຂອງ ຮູບສາມແຈ ທີ່ປະກອບດ້ວຍຂ້າງ a , b , ແລະ c ແມ່ນ
ຮູບສາມແຈທີ່ມີຂ້າງa , b , and c .
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}}
ເຊິ່ງ p ແມ່ນ ເຄິ່ງລວງຮອບ (semiperimeter) ຂອງຮູບສາມແຈ:
p
=
a
+
b
+
c
2
.
{\displaystyle p={\frac {a+b+c}{2}}.}
Heron's formula can also be written as:
S
=
(
a
+
b
+
c
)
(
−
a
+
b
+
c
)
(
a
−
b
+
c
)
(
a
+
b
−
c
)
16
{\displaystyle S={\ {\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\ \over 16}}\,}}
S
=
2
(
a
2
b
2
+
a
2
c
2
+
b
2
c
2
)
−
(
a
4
+
b
4
+
c
4
)
16
{\displaystyle S={\ {\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})\ \over 16}}\,}}
S
=
1
4
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
.
{\displaystyle S={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}.}
ສູດນີ້ແມ່ນຜົນງານຂອງທ່ານ ເຮລົງ ອາເລັກຊານເດຍ, ແລະມີບົດພິສູດຢູ່ໃນປື້ມຂອງລາວຊື່ວ່າ ເມຕຣິກາ Metrica , ຂຽນຂື້ນໃນ ຄ.ສ 60. ເຊິ່ງເຮັດໃຫ້ທ່ານ ອາຄີມີດີດArchimedes ເປັນທີ່ຮູ້ຈັກ, ແລະນັບແຕ່ ປື້ມເມຕຮິກາMetrica ໄດ້ລວບລວມຄວາມຮູ້ກ່ຽວກັບຄະນິດສາດໃນຍຸກບູຮານ,ມັນເປັນໄປໄດ້ວ່ານີ້ເປັນປັດໃຈສຳຄັນໃນການຖືກຳເນີດໃນຜົນງານນີ້. .[ 1]
ສູດ ເຮລົງ ດັ້ງເດີມມີຮູບຮ່າງແມ່ນ:
S
=
1
2
a
2
c
2
−
(
a
2
+
c
2
−
b
2
2
)
2
{\displaystyle S={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}}
a
≥
b
≥
c
{\displaystyle a\geq b\geq c}
ຖືກຄົ້ນພົບໂດຍຄົນຈີນໃນປະເທດເກຣັກ.ແລະຖືກຕີພິມໃນປື້ມຊູຊູ ຈິວຈາງ Shushu Jiuzhang (“Mathematical Treatise in Nine Sections ”), ຂຽນໂດຍ ທ່ານ ຊິນ ຈິວຊາວ Qin Jiushao ພິມໃນປີ ຄ.ສ. 1247.
ການພິສູດລ່າສຸດແມ່ນນຳໃຊ້ພຶດຊະຄະນິດ (algebra) ແລະ ບໍ່ຄ້າຍຄືກັບການພິສູດໂດຍທ່ານເຮລົງເອງ(ໃນໜັງສືເມຕຣິກາ)
ເຮົາຮູ້ວ່າ a , b , c ແມ່ນຂ້າງຂອງຮູບສາມແຈ ແລະ A , B , C ແມ່ນມູມເຊິ່ງໜ້າຂອງຂ້າງດັ່ງກ່າວຕາມລຳດັບ. ເຮົາໄດ້:
cos
C
^
=
a
2
+
b
2
−
c
2
2
a
b
{\displaystyle \cos {\widehat {C}}={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}
ກົດເກນຂອງໂກຊິນ (law of cosines),ຈາກນັ້ນ ເຮົາສາມາດຖອນເອົາຄ່າຂອງ ຊິນ:
sin
C
^
=
1
−
cos
2
C
^
=
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
2
a
b
.
{\displaystyle \sin {\widehat {C}}={\sqrt {1-\cos ^{2}{\widehat {C}}}}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.}
ລວງສູງ ຂອງຮູບສາມແຈ ແມ່ນເທົ່າກັບຜົນຄູນລະຫວ່າງ ຂ້າງພື້ນ a ກັບຂ້າງ b ຄູນໃຫ້ sin(C ),(ອີງຕາມການພົວພັນມູມໃນຮູບສາມແຈ). ເຮົາໄດ້:
S
=
1
2
a
h
=
1
2
a
b
sin
C
^
=
1
4
4
a
2
b
2
−
(
a
2
+
b
2
−
c
2
)
2
=
1
4
(
2
a
b
−
(
a
2
+
b
2
−
c
2
)
)
(
2
a
b
+
(
a
2
+
b
2
−
c
2
)
)
=
1
4
(
c
2
−
(
a
−
b
)
2
)
(
(
a
+
b
)
2
−
c
2
)
=
1
4
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
(
a
+
b
)
−
c
)
(
(
a
+
b
)
+
c
)
=
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
(
a
+
b
)
−
c
)
(
(
a
+
b
)
+
c
)
16
=
(
c
−
(
a
−
b
)
)
2
(
c
+
(
a
−
b
)
)
2
(
(
a
+
b
)
−
c
)
2
(
(
a
+
b
)
+
c
)
2
=
(
b
+
c
−
a
)
2
(
a
+
c
−
b
)
2
(
a
+
b
−
c
)
2
(
a
+
b
+
c
)
2
=
(
a
+
b
+
c
)
2
(
b
+
c
−
a
)
2
(
a
+
c
−
b
)
2
(
a
+
b
−
c
)
2
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
.
{\displaystyle {\begin{aligned}S&={\frac {1}{2}}ah\\&={\frac {1}{2}}ab\sin {\widehat {C}}\\&={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\&={\frac {1}{4}}{\sqrt {(2ab-(a^{2}+b^{2}-c^{2}))(2ab+(a^{2}+b^{2}-c^{2}))}}\\&={\frac {1}{4}}{\sqrt {(c^{2}-(a-b)^{2})((a+b)^{2}-c^{2})}}\\&={\frac {1}{4}}{\sqrt {(c-(a-b))(c+(a-b))((a+b)-c)((a+b)+c)}}\\&={\sqrt {\frac {(c-(a-b))(c+(a-b))((a+b)-c)((a+b)+c)}{16}}}\\&={\sqrt {{\frac {(c-(a-b))}{2}}{\frac {(c+(a-b))}{2}}{\frac {((a+b)-c)}{2}}{\frac {((a+b)+c)}{2}}}}\\&={\sqrt {{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}{\frac {(a+b+c)}{2}}}}\\&={\sqrt {{\frac {(a+b+c)}{2}}{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}}}\\&={\sqrt {p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}.\end{aligned}}}
Proof using the Pythagorean theorem
ດັດແກ້
Triangle with altitude h cutting base c into d + (c − d ). Heron's original proof made use of cyclic quadrilaterals , while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle [2] . The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.
In the form 4T 2 = 4s (s − a )(s − b )(s − c ), Heron's formula reduces on the left to (ch )2 , or
(
c
b
)
2
−
(
c
d
)
2
{\displaystyle (cb)^{2}-(cd)^{2}}
c
2
(
b
2
−
d
2
)
{\displaystyle c^{2}(b^{2}-d^{2})}
using b 2 − d 2 = h 2 by the Pythagorean theorem , and on the right to
(
s
(
s
−
a
)
+
(
s
−
b
)
(
s
−
c
)
)
2
−
(
s
(
s
−
a
)
−
(
s
−
b
)
(
s
−
c
)
)
2
{\displaystyle \displaystyle (s(s-a)+(s-b)(s-c))^{2}-(s(s-a)-(s-b)(s-c))^{2}}
via the identity (p + q ) 2 − (p − q ) 2 = 4pq . It therefore suffices to show
c
b
=
s
(
s
−
a
)
+
(
s
−
b
)
(
s
−
c
)
,
{\displaystyle cb=s(s-a)+(s-b)(s-c),\,}
and
c
d
=
s
(
s
−
a
)
−
(
s
−
b
)
(
s
−
c
)
.
{\displaystyle cd=s(s-a)-(s-b)(s-c).\,}
Then expanding the former you get the following:
2
s
2
−
s
(
a
+
b
+
c
)
+
c
b
.
{\displaystyle 2s^{2}-s(a+b+c)+cb.\,}
and that reduces to
c
b
{\displaystyle cb}
2s = (a + b + c ) and simplifying. Submitting for s the latter s (s − a ) − (s − b )(s − c ) reduces only as far as (b 2 + c 2 − a 2 )/2. But if we replace b 2 by d 2 + h 2 and a 2 by (c − d ) 2 + h 2 , both by Pythagoras, simplification then produces cd as required.
Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative involves arranging the lengths of the sides so that
a
≥
b
≥
c
{\displaystyle a\geq b\geq c}
T
=
1
4
(
a
+
(
b
+
c
)
)
(
c
−
(
a
−
b
)
)
(
c
+
(
a
−
b
)
)
(
a
+
(
b
−
c
)
)
.
{\displaystyle T={\frac {1}{4}}{\sqrt {(a+(b+c))(c-(a-b))(c+(a-b))(a+(b-c))}}.}
The brackets in the above formula are required in order to prevent numerical instability in the evaluation.
Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral . Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral . Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.
Heron's formula is also a special case of the formula of the area of the trapezoid based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero.
Expressing Heron's formula with a Cayley–Menger determinant in terms of the squares of the distances between the three given vertices,
T
=
1
4
−
|
0
a
2
b
2
1
a
2
0
c
2
1
b
2
c
2
0
1
1
1
1
0
|
{\displaystyle T={\frac {1}{4}}{\sqrt {-{\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}}}}}
illustrates its similarity to Tartaglia's formula for the volume of a three-simplex .
Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins .[ 2]
If U , V , W , u , v , w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then[ 3]
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}}}
where
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}}}
↑ ແມ່ແບບ:MathWorld ↑ 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?", [1] , pp. 16-17.
Heath, Thomas L. (1921). A History of Greek Mathematics (Vol II) . Oxford University Press. pp. 321–323. 
