Question

Examine the diagram and information to answer the question.

△ABC has vertices at A(−5,1), B(2,1), and C(6,4).
Point D is located on AC¯¯¯¯¯¯¯¯ in such a way that BD¯¯¯¯¯¯¯¯⊥AC¯¯¯¯¯¯¯¯.
The coordinates of point D are approximately (1.52,2.78).


How many units is the perimeter of △ABC?

29.0 units

26.0 units

23.4 units

20.6 units

Answer 1

23.4 units

Explanation:

Given the point of the vertices of triangle ABC as A(−5,1), B(2,1), and C(6,4).

The distance (d) between two points X (x₁ , y₁) and Y(x₂, y₂) is given as:

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`. We can use this to calculate the length of the sides of the triangle. Therefore:

`AB=√((2-(-5))^2+(1-1)^2)=√(7^2)=7\\ AC=√((6-(-5))^2+(4-1)^2)=√(11^2+3^2)=11.4\\\\BC=√((6-(2))^2+(4-1)^2)=√(4^2+3^2)=5\\`

The perimeter of the triangle ABC = AB + AC + BC = 7 + 11.4 + 5 = 23.4 units