To find the perimeter we need to find the lenght of each segment.
The vertexes of the figure are P(-3,0), Q(-5,3), R(0,3), S(3,5), T(5,2), U(2,0), V(5,-2), W(3,-5), X(0,-3) and Y(-5,-3).
We are going to use the formula:
![d(A,B)=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://img.qammunity.org/2023/formulas/mathematics/high-school/555cxyekk5kg869c1y2u4k9oaarw94dsj1.png)
Now, to simpligy things we notice from the figure (and it is confirmed by the points) that it is symmetric across the x-axis, so we won't need to find all the distances; we are only going to find d(P,Q), d(Q,R), d(R,S), d(S,T), d(T,U), so let's do that:
![\begin{gathered} d(P,Q)=\sqrt[]{(-5-(-3))^2+(3-0)^2} \\ =\sqrt[]{13} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/92krjjlkx57fms55zjeuiy1tzjmlxb28c0.png)
![\begin{gathered} d(Q,R)=\sqrt[]{(0-(-5))^2+(3-3)^2} \\ =5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ugxo8x0tsr98tvt29nrg03etngjnm1jwj8.png)
![\begin{gathered} d(R,S)=\sqrt[]{(3-0)^2+(5-3)^2} \\ =\sqrt[]{13} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qg6vxbfzejjoxdtmxlzo4av9jsyte44xe3.png)
![\begin{gathered} d(S,T)=\sqrt[]{(5-3)^2+(2-5)^2} \\ =\sqrt[]{13} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/46giinlrkc54vs1tfelzo0tpp5kcza2tgx.png)
![\begin{gathered} d(T,U)=\sqrt[]{(2-5)^2+(0-2)^2} \\ =\sqrt[]{13} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u0wiobepb74zvqkh1yx0m1ohqlmqo692u8.png)
With this lengths we can find half of the perimeter, we add them to get:
![5+4\sqrt[]{13}](https://img.qammunity.org/2023/formulas/mathematics/college/yosfojjx5p755p24x7kwxjn4de8mapozie.png)
Hence, the perimeter is:
![\begin{gathered} P=10+8\sqrt[]{13} \\ =38.8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/evcqt6kcgwfb15g7i0ou0lja8y2r61rbd3.png)