Question

The vertices of ABC are A(-2, 2), B(6,2), and CO, 8). The perimeter of ABC?

Answer 1

Answer:

Perimeter = 22.809836694575                              

Step-by-step explanation:

A(-2, 2)  ;  B(6,2)  ;  C(0, 8)

$AB=\sqrt{\left( 6--2\right)^{2} +\left(2-2 \right)^{2} } =\sqrt{64} =8$

$AC=\sqrt{\left( 0--2\right)^{2} +\left(8-2 \right)^{2} } =\sqrt{40} =2\sqrt{10}$

$BC=\sqrt{\left( 0-6\right)^{2} +\left(8-2 \right)^{2} } =\sqrt{72} =6\sqrt{2}$

Then

The perimeter of ΔABC = AB + BC + AC

                                       $=8+6\sqrt{2} +2\sqrt{10}$

                                       = 22.809836694575