what is herons formula?
Answers
Heron's formula, named after Hero of Alexandria, gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.
Answer:
In geometry, Heron's formula (sometimes called Hero's formula), named after Hero of Alexandria,[1] gives the area of a triangle when the length of all three sides are known. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first.
Step-by-step explanation:
Let △ABC be the triangle with sides a = 4, b = 13 and c = 15. This semiperimeter
semiperimeter is s =
1
/
2
(a + b + c) =
1
/
2
(4 + 13 + 15) = 16, and the area is
{\displaystyle {\begin{aligned}A&={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {16\cdot (16-4)\cdot (16-13)\cdot (16-15)}}\\&={\sqrt {16\cdot 12\cdot 3\cdot 1}}={\sqrt {576}}=24.\end{aligned}}}
\begin{align}
A &= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} = \sqrt{16 \cdot (16-4) \cdot (16-13) \cdot (16-15)}\\
&= \sqrt{16 \cdot 12 \cdot 3 \cdot 1} = \sqrt{576} = 24.
\end{align}
In this example, the side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well in cases where one or all of these numbers is not an integer.
History
The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. CE 60. It has been suggested that Archimedes knew the formula over two centuries earlier,[3] and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.[4]
A formula equivalent to Heron's, namely
{\displaystyle A={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}}A=\frac1{2}\sqrt{a^2c^2-\left(\frac{a^2+c^2-b^2}{2}\right)^2}