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Perimeter of A Semicircle: Definition and Examples | EDU.COMEDU.COMResourcesBlogGuidePodcastPlanBackHomesvg]:size-3.5">Math Glossarysvg]:size-3.5">Perimeter of A SemicirclePerimeter of A Semicircle: Definition and ExamplesTable of ContentsPerimeter of a Semicircle Definition of Perimeter of a Semicircle

A semicircle is created when a circle is divided into two equal halves by its diameter. The perimeter of a semicircle is the total length of its boundary, which consists of two parts: half of the circumference of the circle (the curved part) and the length of the diameter (the straight line). Unlike a complete circle where the perimeter is simply the circumference (2πr2\pi r2πr), the perimeter of a semicircle combines both the curved and straight portions.

The formula for calculating the perimeter of a semicircle is πr+2r\pi r + 2rπr+2r or r(π+2)r(\pi + 2)r(π+2), where rrr is the radius of the circle. It's important to note that while a semicircle has half the area of a full circle, its perimeter is not simply half the perimeter of the circle. This is because we need to add the diameter portion that forms the straight boundary of the semicircle.

Examples of Perimeter of Semicircle Calculations Example 1: Finding Perimeter with a Given Radius Problem:

Find the perimeter of the given semicircle with radius 0.70.70.7 inches.

Step-by-step solution:

Step 1, Identify what we know. The radius (r) = 0.70.70.7 inches.

Step 2, Apply the formula for perimeter of a semicircle. We know that perimeter = πr + 2r.

Step 3, Substitute the radius value into the formula. Perimeter = 227×0.7+2×0.7\frac{22}{7} \times 0.7 + 2 \times 0.7722​×0.7+2×0.7

Step 4, Calculate the curved part of the perimeter. 227×0.7=2.2\frac{22}{7} \times 0.7 = 2.2722​×0.7=2.2 inches

Step 5, Calculate the straight part (diameter) of the perimeter. 2×0.7=1.42 \times 0.7 = 1.42×0.7=1.4 inches

Step 6, Add the two parts to find the total perimeter. Perimeter = 2.22.22.2 + 1.41.41.4 = 3.63.63.6 inches

Perimeter of a Semicircle

Example 2: Finding Perimeter with a Given Diameter Problem:

What is the perimeter of a semicircle with diameter 212121 inches?

Step-by-step solution:

Step 1, Convert the diameter to radius. When we know the diameter is 212121 inches, we can find the radius by dividing by 222. Radius (r) = Diameter2=212=10.5\frac{\text{Diameter}}{2} = \frac{21}{2} = 10.52Diameter​=221​=10.5 inches

Step 2, Apply the perimeter formula. We can use the formula: Perimeter = πr + d (where d is the diameter).

Step 3, Substitute the values into the formula. Perimeter = 227×10.5+21\frac{22}{7} \times 10.5 + 21722​×10.5+21

Step 4, Calculate the curved portion of the perimeter. 227×10.5=33\frac{22}{7} \times 10.5 = 33722​×10.5=33 inches

Step 5, Add the curved part and the straight part (diameter) to find the total perimeter. Perimeter = 333333 + 212121 = 545454 inches

Example 3: Finding Radius from Perimeter Problem:

If the perimeter of the semicircle is 7.27.27.2 feet, then find the radius of the circle.

Step-by-step solution:

Step 1, Write what we know. The perimeter of the semicircle is 7.27.27.2 feet.

Step 2, Use the formula for perimeter and set it equal to the given value. Let's use the form r(π+2)r(π + 2)r(π+2) = 7.27.27.2, where r is the radius we're looking for.

Step 3, Substitute π with 227\frac{22}{7}722​ in the equation. r(227+2)=7.2r(\frac{22}{7} + 2) = 7.2r(722​+2)=7.2

Step 4, Simplify the expression inside the parentheses.

r(22+147)=7.2r(\frac{22 + 14}{7}) = 7.2r(722+14​)=7.2 r×367=7.2r \times \frac{36}{7} = 7.2r×736​=7.2

Step 5, Solve for r by multiplying both sides by 736\frac{7}{36}367​.

r=7.2×736r = \frac{7.2 \times 7}{36}r=367.2×7​ r=1.4r = 1.4r=1.4 feet Comments(3)AAdvertiserTobyNovember 6, 2025I've been struggling to explain this to my students. This clear definition and examples really helped! Thanks for the great resource.

FFitnessCoachPeteNovember 5, 2025This glossary page on the perimeter of a semicircle is great! It's helped my students grasp the concept easily with its clear formula and examples.

EEditorHankNovember 4, 2025I've used this to teach my students. The formula and examples made it easy for them to grasp the perimeter of a semicircle. Great resource!

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