Triangle problem.What is the area of the triangle formed by the origin with the pair of points (4,3) and (2,5) ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You also may use Heron's formula to evaluate the area of the triangle, such that:

`A = sqrt(p(p - AB)(p - OA)(p - OB))`

You need to evaluate the lengths of the sides OA,OB,AB, such that:

`OA = sqrt((x_A - x_O)^2 + (y_A - y_O)^2)`

`OA = sqrt((4 - 0)^2 + (3 - 0)^2) `

`OA = sqrt(16 + 9) => OA = 5`

`OB = sqrt(2^2 + 5^2) => OB = sqrt(29)`

`AB = sqrt((2 - 4)^2 + (5 - 3)^2)`

`AB = sqrt(4 + 4) => AB = sqrt 8`

`p = (OA + OB + AB)/2`

`A = sqrt(((OA + OB + AB)/2)((OA + OB - AB)/2)((OB + AB - OA)/2)((OA - OB + AB)/2))`

`A = (1/4)sqrt(((OA + OB)^2 - AB^2)(AB^2 - (OA - OB)^2))`

`A = (1/4)sqrt(((5 + sqrt 29)^2 - 8)(8 - (5 - sqrt29)^2))`

`A = (1/4)sqrt((25 - 29)^2 + 8(5 + sqrt 29)^2 + 8(5 - sqrt29)^2 - 64)`

`A = (1/4)sqrt(8(25 + 10sqrt 29 + 29) + 8(25 - 10sqrt29 + 29) - 48)`

`A = (1/4)sqrt(400 + 464 - 48) => A = (sqrt816)/4`

Hence, evaluating the area of the given triangle, using Heron's formula, yields `A = (sqrt816)/4.`

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