Question

2 Herons
formula
used only to determine
the area of parallelogram

​

Answer 1

Answer:

I think this is write statement

Answer 2

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.

A triangle with sides a, b, and c

Formulation Edit

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

{\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}},}A = \sqrt{s(s-a)(s-b)(s-c)},

where s is the semi-perimeter of the triangle; that is,

{\displaystyle s={\frac {a+b+c}{2}}.}s=\frac{a+b+c}{2}.[2]

Heron's formula can also be written as

{\displaystyle A={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}}A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}

{\displaystyle A={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}}A=\frac{1}{4}\sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}

{\displaystyle A={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}}A=\frac{1}{4}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}

{\displaystyle A={\frac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}}{\displaystyle A={\frac {1}{4}}{\sqrt {4(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{2}+b^{2}+c^{2})^{2}}}}

{\displaystyle A={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}.}A=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}.