Question

What is the perimeter of a triangle with vertices located at (1, 3), (2, 6), and (0, 4), rounded to the nearest hundredth?

Answer 1

The perimeter of a triangle is the sum of the lengths of all sides.
Using the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ calculate the lengths of the sides and then add them up.

$A=(1,3) \\ B=(2,6) \\ C=(0,4) \\ \\ \overline{AB}=\sqrt{(2-1)^2+(6-3)^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10} \\ \overline{AC}=\sqrt{(0-1)^2+(4-3)^2}=\sqrt{(-1)^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \overline{BC}=\sqrt{(0-2)^2+(4-6)^2}=\sqrt{(-2)^2+(-2)^2}=\sqrt{4+4}=\sqrt{4 \times 2}= \\ =2\sqrt{2} \\ \\ P=\overline{AB}+\overline{AC}+\overline{BC}=\sqrt{10}+\sqrt{2}+2\sqrt{2}=\sqrt{10}+3\sqrt{2} \approx 7.40$

The answer is A.

Answer 2

Answer:

7.40 units

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