Question

2. Find the perimeter of the ABCD. (round to the nearest hundredth)

Can anyone please help me with this

Answer 1

The figure is given in the question.

The coordinates are given as

`A(-3,2),B(1,2),C(3,-1),D(0,-3),E(-4,-3)`

Now determine the length AB,

`AB=\sqrt[]{(1-(-3))^2+(2-2)^2}=\sqrt[]{4^2}=4`
`BC=\sqrt[]{(3-1)^2+(-1-2)^2}=\sqrt[]{4+9}=\sqrt[]{13}`
`CD=\sqrt[]{(0-3)^2+(-3-((-1)^2}=\sqrt[]{9+4}=\sqrt[]{13}`
`DE=\sqrt[]{(1-(-3)^2+(2-2)^2}=\sqrt[]{16}=4`

So the perimeter of the figure is the sum of all the sides of the figure.

`AB+BC+CD+DE=4+\sqrt[]{13}+\sqrt[]{13}+4`
`8+2\sqrt[]{13}=15.2`

Hence the perimeter of figure is 15.2.