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

Mathematics

1 answer:

Alexxx [7]1 year ago

3 0

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

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