Question

SOMEONE HELP!!!!!Find the exact perimeter of hexagon ABCDEF plotted below.

Answer 1

Given:

The figure of a hexagon ABCDEF.

To find:

The perimeter of hexagon is ABCDEF.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

From the given figure, it is clear that the vertices of the given figure are A(-6,9), B(5,9), C(8,1), D(2,-6), E(-3,-6) and F(-9,1). We get FA=CB and EF=DC

Using distance formula, we get

$AB=\sqrt{(5-(-6))^2+(9-9)^2}$

$AB=\sqrt{(5+6)^2+(0)^2}$

$AB=\sqrt{(11)^2+0}$

$AB=\sqrt{11}$

Similarly,

$FA=BC=\sqrt{\left(8-5\right)^2+\left(1-9\right)^2}=\sqrt{73}$

$EF=CD=\sqrt{\left(2-8\right)^2+\left(-6-1\right)^2}=\sqrt{85}$

$DE=\sqrt{\left(-3-2\right)^2+\left(-6-\left(-6\right)\right)^2}=5$

Now, the perimeter of the given hexagon is:

$Perimeter=AB+BC+CD+DE+EF+FA$

$Perimeter=11+\sqrt{73}+\sqrt{85}+5+\sqrt{85}+\sqrt{73}$

$Perimeter=16+2\sqrt{73}+2\sqrt{85}$

$Perimeter\approx 51.527$

Therefore, the perimeter of the given hexagon is 51.527.

Answer 2

Answer:

the other person was right, it just seemed wrong because they didn't put the answer we wanted upfront, but its there if you look

Step-by-step explanation:

the picture is proof of the answer