Answer to Question #270003 in Trigonometry for Angel Nodado

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

R (3,0), S (-1,3), and T (0, -2)

$RS=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\sqrt{\left(3-\left(-1\right)\right)^{2}+\left(0-3\right)^{2}}=\sqrt{25}=5$

$RT=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\sqrt{\left(3-0\right)^{2}+\left(0-2\right)^{2}}=\sqrt{13}=3.6$

$ST=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}=\sqrt{\left(\left(-1\right)-0\right)^{2}+\left(3-\left(-2\right)\right)^{2}}=\sqrt{26}=5.1$

So, let's find perimeter now:

$P=RS+RT+ST$

$P=5+3.6+5.1=13.7 units (nearest tenth)$