Question

Write down different forms of Heron's formula.

Answer 1

Heron's formuls - to calculate area of $\Delta$, when height is unkonwn
(0r Heron's formula )
$A=\sqrt{S(S-a)(S-b)(S-c)}S=\frac{a+b+c}{2}=semiperimeter$
$A=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}A=\frac{1}{4}\sqrt{2(a^2+b^2+a^2{c}^2+b^2{c}^2)-(a^4+b^4+c^4)}A=\frac{1}{4}\sqrt{(a^2+b^2+c^2{)}^2-2(a^4+b^4+c^4)}A=\frac{1}{4}\sqrt{4a^2{b}^2-(a^2+b^2-c^2{)}^2}$
Heroin triangle = A $\Delta$ that has side lengths & area tgat are all ingegers.
sometime rational has



$b^2=h^2+d^2$ ---(1)
$a^2=h^2+(c-d{)}^2$ ----(2)
$\left(1\right)-\left(2\right)b^2-a^2=d^2(c-d{)}^2$
$\Rightarrow b^2-a^2((/d)+c-\text{/}d\left)\right(d-c+d)=c(2d-c)$
$\Rightarrow b^2-a^2=2cd-c^2$
$\Rightarrow \frac{c^2+b^2-a^2}{2c}=d\frac{b^2+c^2-a^2}{2c}$\
Now, $h^2=b^2-d^2\Rightarrow h^2=b^2-{\left(\frac{b^2+c^2-a^2}{2c}\right)}^2=\frac{(2bc{)}^2-(b^2+c^2-a^2{)}^2}{4c^2}=\frac{(2bc+b^2+c^2+a^2)(2bc-b^2-c^2+a^2)}{4c^2}=\frac{\left(\right(b+c{)}^2-a^2\left)\right(-(b-c{)}^2+a^2)}{4c^2}\Rightarrow h^2=\frac{\left[\right(b+c{)}^2-a^2\left]\right[a^2-(b-c{)}^2]}{4c^2}=\frac{\left[\right(a+b+c\left)\right(b+c-a\left)\right]\left[\right(a+b-c\left)\right(a-b+c\left)\right]}{4c^2}=\frac{\left[\right(2s\left)2\right(s-a\left)\right]\left[\text{/}2\right(S-c).\text{/}2(S-c).\text{/}2(S-b\left)\right]}{\text{/}4c^2}{h}^2=\frac{4S(S-a)(S-b)(S-c)}{c^2}$
Now, $A=\frac{1}{2}\times c\times h$
$\therefore A=\frac{1}{\text{/}2}\times \mathrm{c/}\times \frac{\text{/}2}{notc}\sqrt{S(S-a)(S-b)(S-c)}$
$\Rightarrow A=\sqrt{S(S-a)(S-b)(S-c)}$
Other formula resembling Heroin's formula


$m_a,m_b,m_c,-lengthofa=\frac{m_a+m_b+m_c}{2}\therefore A=\frac{4}{3}\sqrt{\alpha (\alpha -m_a)(\alpha -m_b)(\alpha -m_c)}$




$h_a,h_b,h_c$ - length of altitudes
H= Semi-sem, of their reciprocars.
$H=\frac{h_a^{-1}+h_b^{-1}+h_c^{-1}}{2}$
$A^{-1}=4\sqrt{H(H-h_a^{-1})(H-h_b^{-1})(H-h_c^{-1})}$