Question

What is the perimeter of a triangle with vertices located at (1,4), (2,7), and (0.5), rounded to the nearest hundreth?

Answer 1

$\boxed {\text {Length = }\sqrt{(X_2-X_1)^2 + (Y_2-Y_1)^2}}$

The three lengths are:

$\text {Length 1 = }\sqrt{(2-1)^2 +(7-4)^2} = \sqrt{10}$

$\text {Length 2 = }\sqrt{(2-0)^2 +(7-5)^2} = \sqrt{8}$

$\text {Length 3 = }\sqrt{(1-0)^2 +(4-5)^2} = \sqrt{2}$

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$\text{Perimeter = } \sqrt{10} + \sqrt{8} + \sqrt{2} = 7.40 \text { (nearest hundredth)}$

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Answer: 7.40 units