Answer:
21.4
Step-by-step explanation:
Given: The coordinates of the vertices of △ABC are A(−2,2), B(5,−3), and C(−4,−1).
To find: Identify the perimeter of △ABC. Round each side length to the nearest tenth.
Solution: To find the perimeter of △ABC, we first need to find the length AB, BC and AC.
We know the distance formula between the coordinates [tex]\left ( x_{1},y_{1} \right ) \text{and} \left ( x_{2},y_{2} \right )[/tex] is [tex]\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
[tex]AB=\sqrt{(5+2)^{2} +(-3-2)^{2} } =\sqrt{49+25}=\sqrt{74}=8.60[/tex]
[tex]BC=\sqrt{(-4-5)^{2} +(-1+3)^{2} } =\sqrt{81+4}=\sqrt{85}=9.21[/tex]
[tex]AC=\sqrt{(-4+2)^{2} +(-1-2)^{2} } =\sqrt{4+9}=\sqrt{13}=3.60[/tex]
Now, rounding each side to the nearest tenth we have,
[tex]AB=8.6[/tex]
[tex]BC=9.2[/tex]
[tex]AC=3.6[/tex]
Now, perimeter of △ABC[tex]=AB+BC+AC[/tex]
[tex]=8.6+9.2+3.6[/tex]
[tex]=21.4[/tex]
Hence, perimeter of △ABC is 21.4.
