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никита6365
@никита6365
April 2021
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Найдите площадь треугольника ABC с вершинами А(11,5) В(3,8) с (6,7)
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lumeire
Тогда
АВ=sqrt[(xB-xA)^2+(yB-yA)^2]=sqrt[(3-11)^2+(8-
5)^2]=sqrt(73); ВC=sqrt[(xC-xB)^2+(yC-yB)^2]=sqrt[(6-3)^2+(7-8)^2]=sqrt(234); АC=sqrt[(xCxA)^2+(yC-yA)^2]=sqrt[(6-11)^2+(-7-5)^2]=sqrt(169)=13.
По теореме косинусов
ВС^2=AB^2+AC^2-2*AB*AC*cos(a) =>cos(a)=(AB^2+AC^2-BC^2)/(2*AB*AC)={[sqrt(73)]^2+[(13)^2[sqrt(234)]^2}/[2*13*sqrt(73)]=8/26*sqrt(73)= 4/13*sqrt(73)sin(a)=sqrt[1-cos^2(a)]=sqrt[1-16/(169*73)]= sqrt[1-16/( 12337)]=sqrt[12321/(12337]=111/13*sqrt(73)
S=(1/2)*AB*AC*sin(a)=(1/2)*sqrt(73)*13*111/13*sqrt(73)=1/2*111=55,5. Ответ: 55,5
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Answers & Comments
Тогда АВ=sqrt[(xB-xA)^2+(yB-yA)^2]=sqrt[(3-11)^2+(8-5)^2]=sqrt(73); ВC=sqrt[(xC-xB)^2+(yC-yB)^2]=sqrt[(6-3)^2+(7-8)^2]=sqrt(234); АC=sqrt[(xCxA)^2+(yC-yA)^2]=sqrt[(6-11)^2+(-7-5)^2]=sqrt(169)=13.
По теореме косинусовВС^2=AB^2+AC^2-2*AB*AC*cos(a) =>cos(a)=(AB^2+AC^2-BC^2)/(2*AB*AC)={[sqrt(73)]^2+[(13)^2[sqrt(234)]^2}/[2*13*sqrt(73)]=8/26*sqrt(73)= 4/13*sqrt(73)sin(a)=sqrt[1-cos^2(a)]=sqrt[1-16/(169*73)]= sqrt[1-16/( 12337)]=sqrt[12321/(12337]=111/13*sqrt(73)
S=(1/2)*AB*AC*sin(a)=(1/2)*sqrt(73)*13*111/13*sqrt(73)=1/2*111=55,5. Ответ: 55,5