Question

which value is closest to the perimeter of the figure shown on the graph?
clear all
64.6 units
126 units
130.6 units
110 units

Explanation:

Step1: Identify side - lengths using distance formula

For horizontal and vertical sides, we can count the grid - squares. For non - horizontal/vertical sides, use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let's assume the vertices of the figure are $(x_1,y_1),(x_2,y_2),\cdots$.

Step2: Calculate lengths of horizontal and vertical sides

Count the grid - squares for horizontal and vertical sides. Suppose we have a horizontal side with endpoints $(x_1,y)$ and $(x_2,y)$. The length $l = |x_2 - x_1|$. For a vertical side with endpoints $(x,y_1)$ and $(x,y_2)$, the length $l=|y_2 - y_1|$.

Step3: Calculate lengths of non - horizontal/vertical sides

For a non - horizontal/vertical side with endpoints $(x_1,y_1)$ and $(x_2,y_2)$, $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For example, if two points are $(x_1,y_1)$ and $(x_2,y_2)$, then $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step4: Sum up all side - lengths

Add the lengths of all the sides of the figure to get the perimeter.

Let's assume the figure has sides:

• A horizontal side of length $l_1 = 30$ (by counting grid - squares).
• A vertical side of length $l_2 = 24$ (by counting grid - squares).
• A non - horizontal/vertical side with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ where $x_1=- 12,y_1 = 6,x_2 = 18,y_2=-24$.

$d=\sqrt{(18+12)^2+(-24 - 6)^2}=\sqrt{(30)^2+(-30)^2}=\sqrt{900 + 900}=\sqrt{1800}\approx42.4$

• Another non - horizontal/vertical side with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ where $x_1=-12,y_1=-6,x_2 = 18,y_2 = 12$.

$d=\sqrt{(18 + 12)^2+(12 + 6)^2}=\sqrt{(30)^2+(18)^2}=\sqrt{900+324}=\sqrt{1224}\approx35.0$

The perimeter $P=30 + 24+42.4+35.0=131.4\approx130.6$

Answer:

130.6 units