A **semicircle** is a half circle. That means a semicircle will have half of the area of a circle. You might think that means it will have half the perimeter of a circle, but that is not true.

To make a semicircle, take any diameter of the circle. Remove one half of the circle along that diameter. You have a semicircle (half of a circle).

A semicircle is half the circumference of a full circle plus the diameter of a cricle, $\left(d\right)$:

Learn about the radius, diameter, and circumference of a circle in this lesson.

The **area** of a semicircle is the space contained by the circle. The area is the number of square units enclosed by the sides of the shape.

The area of a semicircle is always expressed in square units, based on the units used for the radius of a circle.

The formula for the area, $A$, of a circle is built around its radius. You find the area of a semicircle by plugging the given radius of the semicircle into the area of a semicircle formula.

The area formula is:

$A=\frac{\pi {r}^{2}}{2}$

To find the area of a semicircle with diameter, divide the diameter by $2$ to find the radius, and then apply the area of a semicircle formula.

For example, the semicircle below has a radius of $19cm$. What is the area of the semicircle?

To find its area, we replace $r$ with the actual value:

$\mathbf{A}\mathbf{=}\frac{{\mathbf{\pi r}}^{\mathbf{2}}}{\mathbf{2}}$

$A=\frac{\pi \left({19}^{2}\right)}{2}$

$A=\frac{\pi \left(361\right)}{2}$

$A=\frac{1134.114947}{2}$

$A=567.057c{m}^{2}$

The Roman aqueduct of Barcelona in Spain is very old, dating from the first century of the Common Era. The aqueduct is very nearly gone, but it has semicircular arches still visible on a wall in Barcelona.

The arches measure $2.96meters$ in diameter. What is the perimeter and area of each arch?

$\mathbf{P}\mathbf{=}\mathbf{\pi r}\mathbf{+}\mathbf{2}\mathbf{r}$

$P=\pi (1.48m)+2.96m$

$P=4.649557m+2.96m$

$P=7.609557m$

Now, we find the area:

$\mathbf{A}\mathbf{=}\frac{{\mathbf{\pi r}}^{\mathbf{2}}}{\mathbf{2}}$

$A=\frac{\pi \left(1.48{m}^{2}\right)}{2}$

$A=\frac{6.881344{m}^{2}}{2}$

$A=3.440672{m}^{2}$

The **perimeter** of a semicircle is half the original circle's circumference, $C$, *plus* the diameter, $d$. Since the semicircle includes a straight side, its diameter, we cannot describe the distance around the shape as the circumference of a semicircle; it is a perimeter.

Recall that the formula for the perimeter (circumference), $C$, of a circle of radius, $r$, is:

$C=2\pi r$

OR

$C=\pi d$

To find the perimeter, $P$, of a semicircle, you need half of the circle's circumference, *plus* the semicircle's diameter:

$P=\frac{1}{2}\left(2\pi r\right)+d$

The $\frac{1}{2}$ and $2$ cancel each other out, so you can simplify to get this perimeter of a semicircle formula.

$P=\pi r+d$

Using the substitution property of equality, you can also replace diameter with radius throughout:

- $P=\frac{1}{2}\left(2\pi r\right)+2r$
- $P=\pi r+2r$

Let's try an example. A semicircle that has a diameter of $100meters$. What is the perimeter?

$P=\frac{1}{2}\left(\pi d\right)+d$

$P=\frac{1}{2}(\pi \times 100)+100$

$P=\frac{1}{2}\left(314.159265\right)+100$

$P=157.079632+100$

$P=257.08meters$

It is fine to round the decimal places as we did here.

Let's try an example using the radius of a semicircle. A semicircle has a radius of $365inches$. What is its perimeter?

$P=\pi r+2r$

$P=\pi \left(365\right)+2\left(365\right)$

$P=\mathrm{1,146.681318}+730$

$P=\mathrm{1,876.68}inches$

If the question asks you to convert your answer to units like feet or yards, convert it; otherwise leave it in the original linear units. Round your answer to whatever decimal value the problem requires.

The semicircles at both ends of an NBA basketball court indicate the restricted areas beneath each basket. The semicircles have four-foot radii. What is the perimeter of one semicircle in one restricted area?

$P=\pi r+2r$

$P=\pi (4\text{'})+2(4\text{'})$

$P=12.56637\text{'}+8\text{'}$

$P=20.56637\text{'}$

In this case, having a measurement to 100,000ths of a foot is unnecessary; $20.57\text{'}$ is a reasonably accurate answer.

The angle inscribed in a semicircle is always $90\xb0$. The inscribed angle is formed by drawing a line from each end of the diameter to any point on the semicircle. It doesn't matter which point on the length of the arc, the angle created where your two lines meet the arc will always be $90\xb0$.

The two endpoints of the semicircle's diameter and the inscribed angle will always form a right triangle inside the semicircle.

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Get better grades with tutoring from top-rated private tutors. Local and online.

View Tutors

Tutors online

Ashburn, VA

Get better grades with tutoring from top-rated professional tutors. 1-to-1 tailored lessons, flexible scheduling. Get help fast. Want to see the math tutors near you?

Learn faster with a math tutor. Find a tutor locally or online.

Get Started