Calculate the area of triangle ABC with the following vertices A(1, 3, 0) B (0, 2, 5) and C (-1, 0, 2) In radical form.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The formula that contains radical and it gives the area of a triangle is called Heron's formula.

S = sqrt[p(p-a)(p-b)(p-c)]

p = half perimeter of triangle

p = (a+b+c)/2

a,b,c are the lengths of the sides of triangle

a = sqrt[(xC - xB)^2 + (yC - yB)^2]

a = sqrt(1 + 4)

a = sqrt5

b = sqrt[(xC - xA)^2 + (yC - yA)^2]

b = sqrt(4 + 9)

b = sqrt13

c = sqrt[(xA - xB)^2 + (yA - yB)^2]

c = sqrt(1 + 1)

c = sqrt2

S = sqrt{[(sqrt2+sqrt5+sqrt13)/2]*((sqrt2+sqrt5+sqrt13)/2-sqrt5)*((sqrt2+sqrt5+sqrt13)/2-sqrt13)*((sqrt2+sqrt5+sqrt13)/2-sqrt2)]}

S = sqrt{[(sqrt2+sqrt5+sqrt13)/2]*[(sqrt2-sqrt5+sqrt13)/2]*[(sqrt2+sqrt5-sqrt13)/2]*[(-sqrt2+sqrt5+sqrt13)/2]

S = {sqrt[(sqrt2+sqrt5+sqrt13)*(sqrt2-sqrt5+sqrt13)*(sqrt2+sqrt5-sqrt13)*(-sqrt2+sqrt5+sqrt13)]}/4

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