Question

Triangle ABC has vertices at (-3, -1), (-1, 2) and (6, 1). What is the perimeter of triangle ABC, rounded to the nearest tenth?

Answer 1

$Perimeter=19.9units$

Why?

We can find the perimeter of any triangle adding the length of its sides. Since we are given 3 points, we need to calculate the length of the three sides of the triangle,  and then add them to know the perimeter of the triangle.

Let be the points:

A(-3,-1)

B(-1,2)

C(6,1)

- Distances

From A to B:

$d(A,B)=\sqrt{(-1-(-3))^{2}+(2-(-1)^{2}}\\\\d(A,B)=\sqrt{(2)^{2}+(3)^{2}}=\sqrt{4+9}=3.61units$

From A to C:

$d(A,C)=\sqrt{6-(-3))^{2}+(1-(-1)^{2}}\\\\d(A,B)=\sqrt{(9)^{2}+(2)^{2}}=\sqrt{81+4}=9.22units$

From B to C:

$d(B,C)=\sqrt{6-(-1))^{2}+(1-2)^{2}}\\\\d(A,B)=\sqrt{(7)^{2}+(-1)^{2}}=\sqrt{49+1}=7.07units$

Now, calculating the perimeter, we have:

$Perimeter=d(A,B)+d(A,C)+d(B,C)=3.61units+9.22units+7.07units\\\\Perimeter=19.9units$

Have a nice day!