Question

Triangle ABC has vertices at (-4,0) , (-1, 6) and (3,-1) perimeter of triangle ABC, rounded to the nearest tenth ?

Answer 1

Answer:

The perimeter of triangle ABC is $P=21.8\ units$

Step-by-step explanation:

The perimeter of triangle ABC is equal to

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

$A(-4,0),B(-1, 6),C(3,-1)$

step 1

Find the distance AB

$A(-4,0),B(-1, 6)$

substitute in the formula

$d_A_B=\sqrt{(6-0)^{2}+(-1+4)^{2}}$

$d_A_B=\sqrt{(6)^{2}+(3)^{2}}$

$d_A_B=\sqrt{45}\ units$

step 2

Find the distance BC

$B(-1, 6),C(3,-1)$

substitute in the formula

$d_B_C=\sqrt{(-1-6)^{2}+(3+1)^{2}}$

$d_B_C=\sqrt{(-7)^{2}+(4)^{2}}$

$d_B_C=\sqrt{65}\ units$

step 3

Find the distance AC

$A(-4,0),C(3,-1)$

substitute in the formula

$d_A_C=\sqrt{(-1-0)^{2}+(3+4)^{2}}$

$d_A_C=\sqrt{(-1)^{2}+(7)^{2}}$

$d_A_C=\sqrt{50}\ units$

step 4

Find the perimeter

$P=AB+BC+AC$

substitute the values

$P=\sqrt{45}+\sqrt{65}+\sqrt{50}=21.8\ units$