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Find the perimeter of the triangle with vertices at A (0,0), B(-3,-4), and C(-5,0).

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Answer

Perimeter of ABC = 10 + 2√5 = 10 + 4.47 = 14.47

Step-by-step explanation

The perimeter of a figure is given as the sum of the exterior sides of the triangle.

The distance between two points with the coordinates (x₁, y₁) and (x₂, y₂) is given as

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Starting with AB, A (0, 0) and B (-3, -4)

AB = √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(-3 - 0)² + (-4 - 0)²]

AB = √[(-3)² + (-4)²]

AB = √[9 + 16]

AB = √[25] = 5

Going to BC, B (-3, -4) and C (-5, 0)

BC = √[(x₂ - x₁)² + (y₂ - y₁)²]

BC = √[(-5 - (-3))² + (0 - (-4))²]

BC = √[(-5 + 3)² + (0 + 4)²]

BC = √[(-2)² + (4)]

BC = √[4 + 16] = √20 = 2√5

Going to AC, A (0, 0) and C (-5, 0)

AC = √[(x₂ - x₁)² + (y₂ - y₁)²]

AC = √[(-5 - 0)² + (0 - 0)²]

AC = √[(-5)² + (0)²]

AC = √[25 + 0]

AC = √[25] = 5

Perimeter of ABC

= AB + BC + AC

= 5 + 2√5 + 5

= 10 + 2√5

Hope this Helps!!!

User JVXR
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