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 \(\displaystyle 16 + 2\pi\) but I am unsure how they came to it. I'm guessing they get \(\displaystyle 16\) from \(\displaystyle 7 + 3 + 3 + 3\), and \(\displaystyle 2\pi\) by halving the value of the length of the dotted line and multiplying by \(\displaystyle \pi\), but I don't trust myself to guess.

View attachment 8236

 

[MarkFL]

I've edited your post to wrap the expressions containing \pi in [MATH][/MATH] 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:

\(\displaystyle \pi=\frac{C}{D}\implies C=\pi D\implies \frac{C}{2}=\frac{\pi D}{2}\)

And so the perimeter $P$ is:

\(\displaystyle P=3\cdot3+7+\frac{\pi\cdot4}{2}=16+2\pi\)

 

[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 [tex]4\pi[/tex]. The semi-circle has circumference half that, [tex]2\pi[/tex]. The perimeter of the entire figure is [tex]7+ 3+ 3+ 3+ 2\pi= 16+ 2\pi[/tex].