Finding the Perimeter of an Odd Shape.

[RidiculousName]

I have been looking around for information on how to find the perimeter of an odd shape, but I can't find any explanations that seem to fit because the given shape is partially rounded. I already know the answer is $$16 + 2\pi$$ but I am unsure how they came to it. I'm guessing they get $$16$$ from $$7 + 3 + 3 + 3$$, and $$2\pi$$ by halving the value of the length of the dotted line and multiplying by $$\pi$$, but I don't trust myself to guess.

View attachment 8236

 

[MarkFL]

I've edited your post to wrap the expressions containing \pi in $$$$ tags. All $\LaTeX$ markup must be wrapped in tags.

The circular arc is half the circumference of a circle whose diameter is 4 units in length. $\pi$ is defined to be the ratio of the circumference $C$ to the diameter $D$ of a circle:

$$\pi=\frac{C}{D}\implies C=\pi D\implies \frac{C}{2}=\frac{\pi D}{2}$$

And so the perimeter $P$ is:

$$P=3\cdot3+7+\frac{\pi\cdot4}{2}=16+2\pi$$

 

[HOI]

The bottom has length 7. The straight line portion of the top has length 3. That leaves 7- 3= 4 as the diameter of the semi-circle. The circumference of a full circle is "[tex]\pi D[tex], pi times the diameter- here $$4\pi$$. The <b>semi</b>-circle has circumference half that, $$2\pi$$. The perimeter of the entire figure is $$7+ 3+ 3+ 3+ 2\pi= 16+ 2\pi$$.[/tex][/tex]