Question

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

Answer 1

Perimeter = 22.809836694575

Explanation:

A(-2, 2) ; B(6,2) ; C(0, 8)

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

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

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

Then

The perimeter of ΔABC = AB + BC + AC

`=8+6√(2) +2√(10)`

= 22.809836694575