find the perimeter of the triangles whose vertices are (3,10), (5,2), (14,12).
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Hey there,
Perimeter = AB + BC + AC
AB = sqrt[(3 - 5)^2 + (0 - 2)^2]
BC = sqrt[(5 - 7)^2 + (2 - 6)^2]
AC = sqrt[(3 - 7)^2 + (0 - 6)^2]
Perimeter = sqrt[(3 - 5)^2 + (0 - 2)^2] + sqrt[(5 - 7)^2 + (2 - 6)^2] + sqrt[(3 - 7)^2 + (0 - 6)^2]
Perimeter = 14.5 ANSWER
Hope this helps!
Perimeter = AB + BC + AC
AB = sqrt[(3 - 5)^2 + (0 - 2)^2]
BC = sqrt[(5 - 7)^2 + (2 - 6)^2]
AC = sqrt[(3 - 7)^2 + (0 - 6)^2]
Perimeter = sqrt[(3 - 5)^2 + (0 - 2)^2] + sqrt[(5 - 7)^2 + (2 - 6)^2] + sqrt[(3 - 7)^2 + (0 - 6)^2]
Perimeter = 14.5 ANSWER
Hope this helps!
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