MHB Finding the Perimeter of an Odd Shape.

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To find the perimeter of an odd shape that includes a semi-circle, the total perimeter is calculated as the sum of its straight edges and the curved arc. The straight edges contribute lengths of 3, 3, and 7, totaling 16. The semi-circle's diameter is 4, leading to a curved length of 2π, as it is half the circumference of a circle. Therefore, the complete perimeter is expressed as 16 + 2π. Understanding the relationship between the diameter and circumference is crucial for this calculation.
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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.

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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:

$$P=3\cdot3+7+\frac{\pi\cdot4}{2}=16+2\pi$$
 
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 "\pi D, 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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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