Find the area of a triangle whose vertices are A(0,2), B (2,7), C (0,10).
2 answers
I believe the area of the triangle would be 8
AB = sqrt((2-0)^2+(7-2)^2 = 5.39.
BC = sqrt((0-2)^2+(10-7)^2 = 3.61.
AC = sqrt(10-2)^2 = 8.
P = 5.39 + 3.61 + 8 = 17.
A^2 = (P*AB*BC*AC)/16.
a^2 = (17*5.39*3.61*8)/16 = 165.4.
A = sqrt(165.4) = 12.86.
BC = sqrt((0-2)^2+(10-7)^2 = 3.61.
AC = sqrt(10-2)^2 = 8.
P = 5.39 + 3.61 + 8 = 17.
A^2 = (P*AB*BC*AC)/16.
a^2 = (17*5.39*3.61*8)/16 = 165.4.
A = sqrt(165.4) = 12.86.
2 answers