Question

The coordinates of the vertices of △ABC are A(−2,2), B(5,−3), and C(−4,−1). Identify the perimeter of △ABC. Round each side length to the nearest tenth before adding. HELP PLEASE!
Answers:
21.4
30.9
10.7

Answer 1

the distance from (-2,2) to (5,-3) is 8.6,

(5,-3) to (-4,-1) is 9.2

(-4,-1) to (-2,2) is 3.6

8.6+9.2+3.6= 21.4

so the perimeter is 21.4

Answer 2

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 $\left ( x_{1},y_{1} \right ) \text{and} \left ( x_{2},y_{2} \right )$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$AB=\sqrt{(5+2)^{2} +(-3-2)^{2} } =\sqrt{49+25}=\sqrt{74}=8.60$

$BC=\sqrt{(-4-5)^{2} +(-1+3)^{2} } =\sqrt{81+4}=\sqrt{85}=9.21$

$AC=\sqrt{(-4+2)^{2} +(-1-2)^{2} } =\sqrt{4+9}=\sqrt{13}=3.60$

Now, rounding each side to the nearest tenth we have,

$AB=8.6$

$BC=9.2$

$AC=3.6$

Now, perimeter of △ABC$=AB+BC+AC$

$=8.6+9.2+3.6$

$=21.4$

Hence, perimeter of △ABC is 21.4.