Question

Zoe draws ΔABC on the coordinate plane. What is the approximate perimeter of ΔABC

to the nearest hundredth?
A. 8.47 units
B. 12 units
C. 12.94 units
D. 15.31 units

Answer 1

Option C is correct.

Explanation:

Given: Δ ABC on coordinate plane.

From Coordinate plane, Coordinates are A( 1 , 1 ) , B( 3 , 5 ) and C( 5 , 1 )

We need to find Approximate perimeter of the triangle.

we use the distance formula of two point
`(x_1,y_1)\:and\:(x_2,y_2)`

`Distance=√((x_2-x_1)^2+(y_2-y_1)^2)`

So,

`AB=√((3-1)^2+(5-1)^2)=√((2)^2+(4)^2)=√(4+16)=√(20)=4.47\:(approx.)`

`CB=√((5-3)^2+(1-5)^2)=√((2)^2+(-4)^2)=√(4+16)=√(20)=4.47\:(approx.)`

`AC=√((5-1)^2+(1-1)^2)=√((4)^2+(0)^2)=√(16)=4`

Thus, Perimeter = AB + AC + CB = 4.47 + 4.47 + 4 = 12.94 units

Therefore, Option C is correct.