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Find the perimeter of the triangle whose vertices are the following specified (0,3), (-10,-4) (-9,-7) points in the plane.

Sagot :

onyebuchinnaji

Answer:

The perimeter of the triangle is 28.82

Step-by-step explanation:

Please check the image uploaded for the diagram.

Perimeter = d₁ + d₂ + d₃

[tex]d_1 = \sqrt{(-10-0)^2 + (-4-3)^2} = \sqrt{100+49} =\sqrt{149} = 12.21\\\\d_2 = \sqrt{(-9-0)^2 + (-7-3)^2} = \sqrt{81+100} =\sqrt{181} = 13.45\\\\d_3 = \sqrt{(-10+9)^2 + (-4+7)^2} = \sqrt{1+9} =\sqrt{10} = 3.16\\\\Perimeter = 12.21 + 13.45 + 3.16 = 28.82[/tex]

View image onyebuchinnaji
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