Triangle CDE, with vertices C(-8,-7), D(-2,-8) and E(-5,-2), is drawn on the coordinate grid below.
What is the area, in square units, of triangle CDE?


Triangle CDE With Vertices C87 D28 And E52 Is Drawn On The Coordinate Grid Below What Is The Area In Square Units Of Triangle CDE class=

Sagot :

Answer:

Area = 24.75 sqr units

Step-by-step explanation:

You will need these formulas:

[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}[/tex]

Midpoint = [tex](\frac{x_{1} + y_{1} }{2} , \frac{x_{2} + y_{2} }{2})[/tex]

Area = b x h

Let us treat CD as the base. Find the length of the base with the distance formula. Use the coordinates for points C & D.

[tex]d = \sqrt{(-2 - (-8))^2 + (-8-(-7))^2}[/tex]

[tex]d = \sqrt{37}[/tex]

The base is [tex]\sqrt{37}[/tex].

The height is the distance between point E and the midpoint of line CD.

Midpoint of CD = [tex](\frac{-8 + (-7) }{2} , \frac{-2 + (-8) }{2})[/tex] = ([tex]-\frac{15}{2}[/tex], [tex]-5[/tex])

Use the distance formula to find the height.

[tex]d = \sqrt{(-5 - (-\frac{15}{2} ))^2 + (-2-(-5))^2}[/tex]

[tex]d = \frac{\sqrt{61} }{2}[/tex]

Find the area with the two distances that were found.

Area = [tex](\sqrt{37}) (\frac{\sqrt{61} }{2})[/tex]

Area = [tex]\frac{\sqrt{2257} }{2}[/tex]

Area = 24.75 sqr units