Question

question 15 of 40
what is the perimeter of a triangle with vertices located at (-1, 4), (2, 7), and (1, 5), rounded to the nearest hundredth?
a. 7.89 units
b. 9.24 units
c. 7.63 units
d. 8.71 units

Explanation:

Step1: Define distance formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is:
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Step2: Calculate side 1: (-1,4) to (2,7)

Substitute $x_1=-1,y_1=4,x_2=2,y_2=7$
$$d_1=\sqrt{(2-(-1))^2+(7-4)^2}=\sqrt{3^2+3^2}=\sqrt{18}\approx4.2426$$

Step3: Calculate side 2: (2,7) to (1,5)

Substitute $x_1=2,y_1=7,x_2=1,y_2=5$
$$d_2=\sqrt{(1-2)^2+(5-7)^2}=\sqrt{(-1)^2+(-2)^2}=\sqrt{5}\approx2.2361$$

Step4: Calculate side 3: (1,5) to (-1,4)

Substitute $x_1=1,y_1=5,x_2=-1,y_2=4$
$$d_3=\sqrt{(-1-1)^2+(4-5)^2}=\sqrt{(-2)^2+(-1)^2}=\sqrt{5}\approx2.2361$$

Step5: Sum distances for perimeter

$$P=d_1+d_2+d_3\approx4.2426+2.2361+2.2361$$

Answer:

D. 8.71 units