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jessicaa2350
20.05.2020 •
Mathematics
Find the area of the triangle whose vertices are x(3,2) y(-1,0)and z(1,-12)
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Ответ:
The triangle whose vertices are x(3,2) y(-1,0)and z(1,-12).
=> xy = sqrt(4^2 + 2^2) = 4.472
=> xz = sqrt(2^2 + 14^2) = 14.142
=> yz = sqrt(2^2 + 12^2) = 12.166
=> Perimeter/2 = P = (xy + yz + xz)/2 = (4.472 + 14.142 + 12.166)/2 = 15.39
=> Applying Heron's formula, the area of triangle would be:
A = sqrt(P*(P-xy)*(P-xz)*(P-yz))
= sqrt(15.39*(15.39 - 4.472)*(15.39 - 14.142 )*(15.39 - 12.166))
= 26.001
= ~26
Hope this helps!
:)
Ответ: