Triangle ABC has co-ordinates A (–2 ; –2 ), B (1 ; 3) and C (6 ; 0). Determine the length of the legs.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the lengths of the sides of triangle `AB,AC,BC,` using the distance formula, then you need to compare the lengths to decide what are the legs, such that:

`BC = sqrt((x_B - x_C)^2 + (y_B - y_C)^2)`

`BC = sqrt((1 - 6)^2 + (3 - 0)^2)`

`BC = sqrt(25 + 9) => BC = sqrt34`

`AC = sqrt((x_A - x_C)^2 + (y_A - y_C)^2)`

`AC = sqrt((-2 - 6)^2 + (-2 - 0)^2)`

`AC = sqrt68`

`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)`

`AB = sqrt(9 + 25) => AB = sqrt 34`

You should notice that using Pythagora's theorem yields:

`AC^2 = AB^2 + BC^2 => 68 = 34 + 34 => 68 = 68`

Since the triangle ABC is a right isosceles triangle, hence the legs are its shorter sides, thus the legs are AB and BC.

Since the problem requests the lengths of the legs of triangle yields that `AB = BC = sqrt 34` .

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