Question

Help pls o am in 6th grade so take it easy on me pls

Answer 1

From the graph, let's find the perimeter of the figure.

To find the perimeter of the figure, we are to add the length of all sides.

To find the length of each side, let's use the distance formula.

`d=\sqrt[]{(x2-x1)^2+(y2-y1)^2}`

• Length of side with endpoints (-7, 3) and (8, 3):

(x1, y1) ==> (-7, 3)

(x2, y2) ==> (8, 3)

We have:

`\begin{gathered} d=\sqrt[]{(8-(-7))^2+(3-3)^2} \\ \\ d=\sqrt[]{(15)^2+0} \\ \\ d=15 \end{gathered}`

• Length of side with endpoints (8, 3) and (8, -8):

(x1, y1) ==> (8, 3)

(x2, y2) ==> (8, -8)

We have:

`\begin{gathered} d=\sqrt[]{(8-8)^2+(-8-3)^2} \\ \\ d=\sqrt[]{11^2} \\ \\ d=11 \end{gathered}`

• Length of side with endpointd (2, -1) and (2, -8):

(x1, y1) ==> (2, -1)

(x2, y2) ==> (2, -8)

We have:

`\begin{gathered} d=\sqrt[]{(2-2)^2+(-8-(-1))^2} \\ \\ d=\sqrt[]{(-7)^2} \\ \\ d=7 \end{gathered}`

• Length of sides with endpoints (2, -8) and (8, -8):

(x1, y1) ==> (2, -8)

(x2, y2) ==> (8, -8)

We have:

`\begin{gathered} d=\sqrt[]{(8-2)^2+(-8-(-8))^2} \\ \\ d=\sqrt[]{6^2} \\ \\ d=6 \end{gathered}`

• Length of side with endpoints (-7, -1) and (2, -1):

(x1, y1) ==> (-7, -1)

(x2, y1) ==> (2, -1)

We have:

`\begin{gathered} d=\sqrt[]{(2-(-7))^2+(-1-(-1))^2} \\ \\ d=\sqrt[]{9^2} \\ \\ d=9 \end{gathered}`

Length of side with endpoint (7, 3) and (-7, -1):

(x1, y1) ==> (-7, 3)

(x2, y2) ==> (-7, -1)

We have:

`\begin{gathered} d=\sqrt[]{(-7-(-7))^2+(-1-3)^2} \\ \\ d=\sqrt[]{(-4)^2} \\ \\ d=4 \end{gathered}`

Now, we have the length of all 6 sides of the figure.

To find the perimeter of the figure, let's sum up all 6 sides.

Perimeter = 15 + 11 + 7 + 6 + 9 + 4

Perimeter = 52 units

Therefore, the perimeter of the given figure is 52 units.

ANSWER:

52 units