Answer to Question #270790 in Geometry for kram

Question

The vertices of a triangle are R (3,0), S (-1,3), and T (0, -2). a) Classify the triangle by side length and b) Find the perimeter of the triangle to the nearest tenth.

Answer 1

a)

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

$$RT=\sqrt{(0-3)^2+(-2-0)^2}=\sqrt{13}$$

$$TS=\sqrt{(-1-0)^2+(-2-3)^2}=\sqrt{26}$$

It is a versatile triangle.

b)

$$P_{RST}=5+\sqrt{13}+\sqrt{26}\approx13.7$$