A=(4,0) B=(-2,-5), C=(0,2)
ze wzoru na długość odcinka w układzie współrzędnych:
|AB|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}
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a=|AB|=\sqrt{(-2+4)^2+(-5+0)^2}=\sqrt{2^2+(-5)^2}=\sqrt{4+25}=\sqrt{29}
b=|BC|=\sqrt{(0+2)^2+(2+5)^2}=\sqrt{2^2+7^2}=\sqrt{4+49}=\sqrt{53}
c=|AC|=\sqrt{(0-4)^2+(2-0)^2}=\sqrt{(-4)^2+2^2}=\sqrt{16+4}=\sqrt{20}
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p=\frac{a+b+c}{2}
p=\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2} połowa obwodu
P=p(p-a)(p-b)(p-c) wzór Herona
P=\sqrt{\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2}*(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2}-\sqrt{29})(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2}-\sqrt{53})(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2}-\sqrt{20})}=
=\sqrt{\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}}{2}*(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}-2\sqrt{29}}{2})(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}-2\sqrt{53}}{2})(\frac{\sqrt{29}+\sqrt{53}+\sqrt{20}-2\sqrt{20}}{2})}=
=\sqrt{\frac{\sqrt{53}+\sqrt{20}+\sqrt{29}}{2}*(\frac{\sqrt{53}+\sqrt{20}-\sqrt{29}}{2}*\frac{\sqrt{29}+\sqrt{20}-\sqrt{53}}{2}*\frac{\sqrt{29}+\sqrt{53}-\sqrt{20}}{2}}=
=\sqrt{\frac{(\sqrt{53}+\sqrt{20}+\sqrt{29})(\sqrt{53}+\sqrt{20}-\sqrt{29})}{4}*\frac{(\sqrt{29}+\sqrt{20}-\sqrt{53})(\sqrt{29}+\sqrt{53}-\sqrt{20})}{4}}=
=\sqrt{\frac{(\sqrt{53}+\sqrt{20})^2-(\sqrt{29})^2}{4}*\frac{29+\sqrt{29*53}-\sqrt{20*29}-\sqrt{29*53}-53*\sqrt{53*20}+\sqrt{20*29}+\sqrt{20*53}-20}{4}}=
=\sqrt{\frac{53*2\sqrt{53*20}+20-29}{4}*\frac{2*\sqrt{53*20}-44}{4}}=
=\sqrt{\frac{(2\sqrt{1060}+44)*(2\sqrt{1060}-44)}{16}}=
=\sqrt{\frac{(2\sqrt{4*265}+44)*(2\sqrt{4*265}-44)}{16}}=
=\sqrt{\frac{(4\sqrt{265}+44)*(4\sqrt{265}-44)}{16}}=
=\sqrt{\frac{\not4^1(\sqrt{265}+11)*\not4^1(\sqrt{265}-11)}{\not4^1*\not4^1}}=
=\sqrt{(\sqrt{265}+11)(\sqrt{265}-11)}=
=\sqrt{\frac{(\sqrt{265})^2-11^2}{1}}=
=\sqrt{265-121}=\sqrt{144}=12