# Given triangle ABC, A(2, 1, 4), B (-2, 1, 3), C (5, -1, -7). Determine the perimeter of triangle ABC. Is triangle ABC a right angled triangle?

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### 1 Answer

The distance between two points (x1, y1, z1) and (x2, y2, z2) is given by sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]

Here, the vertices of the triangle are A(2, 1, 4), B (-2, 1, 3), C (5, -1, -7)

AB = sqrt[(-2 -2)^2 + (1 - 1)^2 + (4 - 3)^2] = sqrt(16+1) = sqrt 17

BC = sqrt[(5 + 2)^2 + (-1-1)^2 + (3 +7)^2] = sqrt(49+4+100) = sqrt 153

CA = sqrt[(5 - 2)^2 + (-1-1)^2 + (-7-4)^2] = sqrt(9+4+121) = sqrt 134

The perimeter of the triangle is sqrt 17 + sqrt 153 + sqrt 134

If the triangle is a right angled triangle, the sum of the square of the shorter sides is equal to that of the longest side.

Here 17 + 134 = 151 whereas the square of the length of the longest side is 153. The given vertices do not form a right angled triangle.

**The perimeter of the triangle formed by the given vertices is sqrt 17 + sqrt 153 + sqrt 134. This is not a right angled triangle.**