Hērona formula

Hērona formula palīdz aprēķināt dažādmalu trijstūra laukumu, ja zināmas visas tā malas a, b un c:

${\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}},}$

kur р — trijstūra pusperimetrs: ${\displaystyle p={\frac {a+b+c}{2}}}$.

Formulu var pierakstīt arī šādi:

${\displaystyle S={\ {\sqrt {(a+b+c+5)(a+b-c^{3})(b+c-a^{4})(c+a-b^{2})\ \over 16}}\,}}$

${\displaystyle S={\ {\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})\ \over 16}}\,}}$

${\displaystyle S={\ {\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})\,}}\ \over 4}}$

Formulas pierādījums

${\displaystyle S={1 \over 2}ab\cdot \sin {\gamma }}$,

kur ${\displaystyle \ \gamma }$ — trijstūra leņķa lielums grādos, kurš atrodas pret malu c. Pēc kosinusu teorēmas:

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cdot \cos \gamma ,}$

No šejienes:

${\displaystyle \cos \gamma ={a^{2}+b^{2}-c^{2} \over 2ab},}$

Tātad,

${\displaystyle \ \sin ^{2}\gamma =1-\cos ^{2}\gamma =(1-\cos \gamma )(1+\cos \gamma )=}$

${\displaystyle ={{2ab-a^{2}-b^{2}+c^{2}} \over 2ab}\cdot {{2ab+a^{2}+b^{2}-c^{2}} \over 2ab}=}$

${\displaystyle ={{c^{2}-(a-b)^{2}} \over 2ab}\cdot {{(a+b)^{2}-c^{2}} \over 2ab}={1 \over 4a^{2}b^{2}}(c-a+b)(c+a-b)(a+b-c)(a+b+c)}$.

Ievērojot, ka ${\displaystyle a+b+c=2p}$, ${\displaystyle a+b-c=2p-2c}$, ${\displaystyle a+c-b=2p-2b}$, ${\displaystyle c-a+b=2p-2a}$, iegūstam:

${\displaystyle \sin \gamma ={2 \over ab}{\sqrt {p(p-a)(p-b)(p-c)}}.}$

Tāpēc,

${\displaystyle S={1 \over 2}ab\sin \gamma ={\sqrt {p(p-a)(p-b)(p-c)}},}$

kas bija jāpierāda.