Question

the verticales of triangle XYZ are X(-3,-6),Y(21,-6),and Z(21,4). What is the perimeter of the triangle​

Answer 1

Answer:

P = 60

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

X(-3, -6), Y(21, -6). Substitute:

$XY=\sqrt{21-(-3))^2+(-6-(-6))^2}=\sqrt{24^2+0^2}=\sqrt{24^2}=24$

X(-3, -6), Z(21, 4). Substitute:

$XZ=\sqrt{(21-(-3))^2+(4-(-6))^2}=\sqrt{24^2+10^2}=\sqrt{576+100}=\sqrt{676}=26$

Y(21, -6), Z(21, 4). Substitute:

$YZ=\sqrt{(21-21)^2+(4-(-6))^2}=\sqrt{0^2+10^2}=\sqrt{10^2}=10$

The perimeter of the triangle XYZ:

$P_{\triangle XYZ}=XY+XZ+YZ$

Substitute:

$P_{\triangle XYZ}=24+26+10=60$