From the diagram, we can see that $\triangle{DBC}$ is an isosceles triangle with $BD = DC$ and $\triangle{ABC}$ is an equilateral triangle.

We're told that the perimeter of $\triangle ABC$ is $p_1=24,$ and since all $3$ sides have equal length, each side must have a length of \[ \dfrac{p_1}{3}= \dfrac{24}{3} = 8. \]

Now we can find $BD\mathbin{:}$ \[ BD = \dfrac{3}{2}{AB} =\dfrac{3}{2}(8)=12 \]

So, we know that $AB=BC=AC = 8,$ and $BD = DC = 12.$ Hence, we have enough information to compute the perimeter of the triangle $\triangle{CBD}\mathbin{:}$ \begin{align*} p &= CB +BD + DC \\[2pt] &=8 + 12 + 12 \\[2pt] &=32 \end{align*}

From the diagram, we can see that $\triangle{ABC}$ is an equilateral triangle and $\triangle{DBC}$ is an isosceles triangle with $CD=DB.$

We're told that the perimeter of $\triangle DBC$ is $p_1=10,$ and $DB=3,$ so we get \begin{align*} p_1 &= DB +BC + CD \\[2pt] 10 &=3+BC+3 \\[2pt] 4 &=BC. \end{align*}

So now we have enough information to compute the perimeter of the triangle $\triangle{ABC}\mathbin{:}$ \begin{align*} p &= 3 BC\\[2pt] &=3\cdot4\\[2pt] &=12 \end{align*}

Therefore, the perimeter of $\triangle ABC$ is $p=12.$

The perimeter $p$ of a polygon is the sum of the measures of all its sides. So, we can simply calculate \begin{align*} p &= AB + BC + CD + DE + EA\\ &= 11 + 3 + 12 + 4 + 20\\ &=50\,. \end{align*}

The perimeter $p$ of a polygon is the sum of the measures of all its sides. Since all the sides of a regular hexagon are equal and the measure of one of them is $4\, \mathrm{cm}$, we have \[ p = 6\cdot AB = 6\cdot 4 = 24\, \mathrm{cm}. \]

The perimeter of the octagon shown below is $10$, as well as the perimeter of the given triangle. Among the options given, the octagon is the only one that satisfies this.

Since all the sides of a square are equal, the perimeter of the square is

\[p=4\cdot 8=32.\]

We know that our polygons have equal perimeters. Therefore, \begin{align*} 32&= \overbrace{{AB} +{BC} +{CD} +{DE} +{EF} + FG + GH + HA}^{\large\textrm{Perimeter of $ABCDEFGH$}} \\[2pt] 32&= 8x \\[2pt] 4 &=x. \end{align*}

So, $x=4.$