Question

Find the perimeter of the triangle whose vertices are the following specified points in the plane. (4,−7),(−10,−4) and (3,−7)

Answer 1

Answer:

  √205 +√178 +1 ≈ 28.6595

Step-by-step explanation:

The distance formula tells you the distance between points (x1, y1) and (x2, y2) is ...

  d = √((x2 -x1)^2 +(y2-y1)^2)

Taken pairwise, the distances between the given points are ...

  d1 = √((-10-4)^2 +(-4-(-7))^2) = √(196+9) = √205

  d2 = √((3-(-10))^2 +(-7-(-4))^2) = √(169+9) = √178

  d3 = √((4-3)^2 +(-7-(-7))^2) = √1 = 1

Then the sum of the distances gives the perimeter:

  P = d1 +d2 +d3 = √205 +√178 +1 ≈ 28.6595

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In the attached figure, the distance "a" is the sum of the two long sides of the triangle.