what is formula of herons formula?
Answers
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Answer:
Heron's formula is a formula that can be used to find the area of a triangle, when given its three side lengths. It can be applied to any shape of triangle, as long as we know its three side lengths. The formula is as follows:
Explanation:
You can calculate the area of a triangle if you know the lengths of all three sides, using a formula that has been known for nearly 2000 years. It is called "Heron's Formula" after Hero of Alexandria (see below) Just use this two step process: Step 1: Calculate "s" (half of the triangles perimeter): s = a+b+c2.
The area of a triangle whose side lengths are a, b,a,b, and cc is given by
A=\sqrt{s(s-a)(s-b)(s-c)},
A=
s(s−a)(s−b)(s−c)
,
where s=\dfrac{(\text{perimeter of the triangle})}{2}=\dfrac{a+b+c}{2}s=
2
(perimeter of the triangle)
=
2
a+b+c
, semi-perimeter of the triangle.
Other useful forms are
\begin{aligned} A&=\frac 1 4\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\ \\ A&=\frac 1 4\sqrt{ \big[(a+b+c)(a+b-c) \big] \times \Big[\big(+(a-b)+c\big)\big(-(a-b)+c\big) \Big]}\\ A&=\frac 1 4\sqrt{\Big[(a+b)^2-c^2\Big] \times \ \Big[c^2-(a-b)^2\Big] }\\ \\ A&=\frac{1}{4}\sqrt{4a^2b^2-\big(a^2+b^2-c^2\big)^2}\\ A&=\frac{1}{4}\sqrt{2\left(a^2 b^2+a^2c^2+b^2c^2\right)-\left(a^4+b^4+c^4\right)} \\ A&=\frac{1}{4}\sqrt{\left(a^2+b^2+c^2\right)^2-2\left(a^4+b^4+c^4\right)}. \end{aligned}
A
A
A
A
A
A
=
4
1
(a+b+c)(a+b−c)(a−b+c)(−a+b+c)
=
4
1
[(a+b+c)(a+b−c)]×[(+(a−b)+c)(−(a−b)+c)]
=
4
1
[(a+b)
2
−c
2
]× [c
2
−(a−b)
2
]
=
4
1
4a
2
b
2
−(a
2
+b
2
−c
2
)
2
=
4
1
2(a
2
b
2
+a
2
c
2
+b
2
c
2
)−(a
4
+b
4
+c
4
)
=
4
1
(a
2
+b
2
+c
2
)
2
−2(a
4
+b
4
+c
4
)