TriangleABC has vertices A(10,8) B(-2,10) and C(2,4). Classify the triangle.

Question

Determine the perimeter of the triangle?

Answer 1

The vertices are A(10, 8), B(-2, 10) and C(2, 4).

Length of the side AB = √[(10 - (-2))² + (8 - 10)²] = √[(12)² + (-2)²] = √[144 + 4] = √148.

Length of the side BC = √[(-2 - 2)² + (10 - 4)²] = √[(-4)² + (6)²] = √[16 + 36] = √52.

Length of the side AC = √[(10 - 2)² + (8 - 4)²] = √[(8)² + (4)²] = √[64 + 16] = √80.

Perimeter of the triangle = AB + BC + AC

= √148 + √52 + √80

= √[4(37)] + √[4(13)] + √[16(5)]

= 2√(37) + 2√(13) + 4√(5)

= 2(6.08) + 2(3.6) + 4(2.23)

= 12.16 + 7.2 + 8.94

= 28.3 units.