Question

Zoe draws ABC on the coordinate plane.

ТУ
B
5
4
3
2.
1
А
ol
1
12
13
4
5
What is the approximate perimeter of AABC to the nearest hundredth?
O A 8.47 units
o
B
12 units
C 12.94 units
O D. 15.31 units

Answer 1

12.94 units

Explanation:

Perimeter of ∆ABC = AB + BC + AC

✔️Distance between A(1, 1) and B(3, 5):

`AB = √((x_2 - x_1)^2 + (y_2 - y_1)^2)`

Where,

`A(1, 1) = (x_1, y_1)`

`B(3, 5) = (x_2, y_2)`

`AB = √((3 - 1)^2 + (5 - 1)^2)`

`AB = √((2)^2 + (4)^2)`

`AB = √(4 + 16)`

`AB = √(20)`

AB = 4.47 units

✔️Distance between B(3, 5) and C(5, 1)

`BC = √((x_2 - x_1)^2 + (y_2 - y_1)^2)`

Where,

`B(3, 5) = (x_1, y_1)`

`C(5, 1) = (x_2, y_2)`

`BC = √((5 - 3)^2 + (1 - 5)^2)`

`BC = √((2)^2 + (-4)^2)`

`BC = √(4 + 16)`

`BC = √(20)`

BC = 4.47 units

✔️Distance between A(1, 1) and C(5, 1):

AC = |1 - 5| = 4 units

✅Perimeter of ∆ABC = 4.47 + 4.47 + 4 = 12.94 units