Question

The coordinates of the verticies of triangle ABC are A(2,5), B(6,-1), and C(-4,-2). Fint the perimeter of triangle ABC, to the nearest tenth

Answer 1

Answer:

24.4 Units

Step-by-step explanation:

For this, we will use the distance formula to find the distance between each vertex

$\sqrt({x_{1}-x_{2} })^{2} +(y_{1} -y_{2})^{2}$

So we do this for each side of the triangle to get AB = $\sqrt{52}$, BC = $\sqrt{109}$, and AC = $\sqrt{45}$

We add these all up to get 24.4 (rounded)