Q:

Unit 4 midFind the Perimeter of the triangle. Round 2 decimal places.

Accepted Solution

A:
A (-2, 3) B(3, 0) C(-4,-3)

[tex]\text {Length of AB = } \sqrt{(-2-3)^2+ (3-0)^2} [/tex]

[tex]\text {Length of AB = } \sqrt{(-5)^2+ (3)^2} [/tex]

[tex]\text {Length of AB = } \sqrt{25 + 9} [/tex]

[tex]\text {Length of AB = } \sqrt{34} [/tex]

[tex]\text {Length of AB = } 5.83 [/tex]

[tex]\text {Length of BC = } \sqrt{(3+4)^2+ (0+3)^2}[/tex]

[tex]\text {Length of BC = } \sqrt{(7)^2+ (3)^2}[/tex]

[tex]\text {Length of BC = } \sqrt{49 + 9}[/tex]

[tex]\text {Length of BC = } \sqrt{58}[/tex]

[tex]\text {Length of BC = } 7.62[/tex]

[tex]\text {Length of AC = } \sqrt{(-2+4)^2+ (3+3)^2}[/tex]

[tex]\text {Length of AC = } \sqrt{(2)^2+ (6)^2}[/tex]

[tex]\text {Length of AC = } \sqrt{4 + 12}[/tex]

[tex]\text {Length of AC = } \sqrt{16}[/tex]

[tex]\text {Length of AC = } 4}[/tex]

Perimeter = 5.83 +  7.62 + 4 = 17.45 units