Aptarimas:Matematika/Trikampiai

Trikampio ploto radimas žinant koordinates

Vaizdas:Trapecijaris11-4.jpg

11.

Trikampio plotas. Bet kokiems taškams $A(x_{1};y_{1})$, $B(x_{2};y_{2})$ ir $C(x_{3};y_{3})$ negulintiems ant vienos tiesės, plotas S trikampio ABC išreiškiamas formule

$S={\frac {1}{2}}|[(x_{2}-x_{1})(y_{3}-y_{1})-(x_{3}-x_{1})(y_{2}-y_{1})]|.$

Įrodymas. Plotą trikampio ABC pavaizduotą pav. 11, galima rasti taip:

$S_{ABC}=S_{ADEC}+S_{BCEF}-S_{ABFD},\;$

kur $S_{ADEC}$, $S_{BCEF}$, $S_{ABFD}$ - plotai atitinkamų trapecijų.

Kadangi

$S_{ADEC}=|DE|{\frac {|AD|+|CE|}{2}}={\frac {(x_{3}-x_{1})(y_{3}+y_{1})}{2}},$

$S_{BCEF}=|EF|{\frac {|EC|+|BF|}{2}}={\frac {(x_{2}-x_{3})(y_{2}+y_{3})}{2}},$

$S_{ABFD}=|DF|{\frac {|AD|+|BF|}{2}}={\frac {(x_{2}-x_{1})(y_{1}+y_{2})}{2}},$

įstatę išraiškas šiems plotams į lygybę $S_{ABC}=S_{ADEC}+S_{BCEF}-S_{ABFD},$ gausime formulę

$S_{ABC}=S_{ADEC}+S_{BCEF}-S_{ABFD}={\frac {1}{2}}|[(x_{3}-x_{1})(y_{3}+y_{1})+(x_{2}-x_{3})(y_{2}+y_{3})-(x_{2}-x_{1})(y_{1}+y_{2})]|=$

$={\frac {1}{2}}|[(x_{1}-x_{2})(y_{1}+y_{2})+(x_{2}-x_{3})(y_{2}+y_{3})+(x_{3}-x_{1})(y_{3}+y_{1})]|=$

$={\frac {1}{2}}|(x_{1}y_{1}+x_{1}y_{2}-x_{2}y_{1}-x_{2}y_{2}+x_{2}y_{2}+x_{2}y_{3}-x_{3}y_{2}-x_{3}y_{3}+x_{3}y_{3}+x_{3}y_{1}-x_{1}y_{3}-x_{1}y_{1})|=$

$={\frac {1}{2}}|(x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3})|={\frac {1}{2}}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|=$

$={\frac {1}{2}}|(x_{2}-x_{1})(y_{3}-y_{1})-(x_{3}-x_{1})(y_{2}-y_{1})|.$

Pavyzdis. Duoti taškai A(1; 1), B(6; 4), C(8; 2). Rasti trikampio ABC plotą. Randame:

$S_{ABC}={\frac {1}{2}}|[(x_{2}-x_{1})(y_{3}-y_{1})-(x_{3}-x_{1})(y_{2}-y_{1})]|={\frac {1}{2}}|(6-1)(2-1)-(8-1)(4-1)|={\frac {1}{2}}|5\cdot 1-7\cdot 3|={\frac {1}{2}}|5-21|={\frac {1}{2}}|-16|={\frac {16}{2}}=8;$

$S_{ABC}=S_{ADEC}+S_{BCEF}-S_{ABFD}={\frac {1}{2}}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|={\frac {1}{2}}|1(4-2)+6(2-1)+8(1-4)|={\frac {1}{2}}|2+6+8\cdot (-3)|=$

$={\frac {1}{2}}|8-24|={\frac {1}{2}}|-16|=8.$