## Introduction

As we explore more geometry concepts, we encounter another two-dimensional shape called a semi-circle. From a taco inside the restaurant to the moon during a half-moon, we observe these things to have a semi-circular shape.

In this article, we learn the fundamental concepts starting with definitions of the shape, then we quantify the properties of the semi-circle by defining both its perimeter and then its area. While doing so, we will also try some examples to guide us in learning.

**What is a Semi-Circle?**

A **semi-circle**, by definition, is **half of a circle**. We can think of a circle being sliced into two halves, and both shapes turn out to be semi-circles. To visualize, here is what a semi-circle looks like:

In geometric terms, a semi-circle is formed by an **arc** covering half of a circle, and a straight line passing through the center connecting both ends of the arc called its **diameter**.

Like a circle, any point in the arc of the semicircle is **equidistant **from its center. A line connecting the center to any point in the arc of the semicircle has a length equal to the **radius **of the semicircle.

**What is a Perimeter?**

We define the **perimeter** of a two-dimensional shape as a **measure of the boundary enclosing it**.

In an intuitive approach, we can relate the idea of a perimeter to how far we jog through a lap in the park. If we begin measuring the distance from the start of the lap, and then trace a closed path until we get back to the starting point, we can measure the perimeter of the park through the lap we have ran across:

Thus, for any closed shape we can get its perimeter by taking the length of the edges enclosing the shape.

**What is an Area?**

On the other hand, we can define the **area **of a two-dimensional shape as a **measure of the space it occupies in the two-dimensional plane**.

Intuitively, we can think of shapes in a plane as objects in a room. Each object takes up some space in the room, depending on the kind of object and its size. In this sense, we can measure a shape’s area through the shape’s dimensions.

**What is Pi?**

Before we proceed with computing for the perimeter and area of a semi-circle, we also introduce a special number that will prove relevant to our computations later.

The mathematical constant **pi**, denoted by , is a number that relates the diameter of a circle to its circumference. If we measure the perimeter of a circle and divide it with the measure of its diameter, we get the approximate value of pi:

π≈3.1415926…

It is important to note that this constant is a **non-terminating irrational number**. This means that the exact decimal value of pi is a never-ending stream of numbers! However, for most cases we only round off the value of to the nearest hundredths:

π≈3.14

## Perimeter of a Semi-Circle

From the definitions provided earlier, we can now discuss in detail the formula for solving the perimeter of a semi-circle. We first derive the basic formula involved, provide intuitive insights, and then show how we apply this formula with some examples.

**Derivation of the Perimeter Formula of a Semi-Circle**

We know that the perimeter can be defined as the sum of the lengths of all edges covering a shape. In a semi-circle, we have two edges: a curved arc and a diameter. Hence, we can say that the perimeter of a semi-circle can be expressed as the sum of the length of the arc and the length of the diameter of the semi-circle:

Perimeter=Length of Arc+Length of Diameter

Next, we know that the length of the arc spans half of a circle. In other words, we can say that the length of the arc is half of the circumference of a circle:

Length of Arc=½ x 2πr

∴Length of Arc=πr

Then, we can describe the diameter of a semi-circle as twice the length of its radius:

Length of Diameter=2×r

∴Length of Diameter=2r

Combining these expressions, we can now express the perimeter as a sum in terms of the radius r:

Perimeter=πr+2r

Finally, grouping the terms we arrive at the formula for the perimeter of a semi-circle:

Perimeter=r×(π+2)

In case the diameter d is known and the radius is not known to us, we can rewrite the radius as half of the diameter:

r=$\frac{d}{2}$

Substituting this expression into the formula for the perimeter, we obtain an alternative formula for the perimeter of a semi-circle using the diameter:

Perimeter=$\frac{d}{2}$×(π+2)

**What is the Perimeter Formula of a Semi-Circle?**

From the previous derivation performed, the perimeter P of a semi-circle with a known radius r is given by:

P=r×(π+2)

If the diameter d of the semi-circle is given instead of a radius, we use an alternative formula that is expressed in terms of the diameter:

P=$\frac{d}{2}$×(π+2)

We note that for both formulas, the **unit of the perimeter** is expressed in terms of the **same units** as with the radius/diameter given.

**How is it Used?**

As a practice on how to apply the formula, let us work together on finding the perimeter of the semi-circle shown below:

From the given figure, we are given the radius of the semi-circle to be three units. Hence, we can say that:

r=3 units

Moreover, since the radius is known, we can use the Perimeter Formula in terms of the radius r:

P=r×(π+2)

We then substitute the given radius to the formula, along with an approximate value of π≈3.14 to get:

P=3 units×(3.14+2)

Afterward, we add the numbers within the grouping symbols:

P=3 units×(5.14)

Then, multiplying the given radius by the sum obtained earlier we obtain the perimeter of the semi-circle:

P=15.42 units

We then conclude that the perimeter of the semi-circle is 15.42 units. This quantity is expressed in terms of the same units as the radius.

## Area of a Semi-Circle

Following the definition of a semi-circle, we can find similarities in the computation of the area of a semi-circle based on the area of a circle it forms.

**Derivation of the Area Formula of a Semi-Circle**

We first recall the Area Formula of a circle whose radius r is known:

A_{circle}=πr^{2}

If a semi-circle is half of a circle, then the space it would occupy would be half of the space a circle occupies. Hence, we can say that the area of a semi-circle is half of the area of a circle:

Area=½ x A_{circle}

If we substitute the Area Formula for the circle, we then arrive at the Area Formula for a semi-circle:

Area=½ πr^{2}

In case the diameter d is known and the radius is not known to us, we can rewrite the radius as half of the diameter:

r=$\frac{d}{2}$

Substituting this expression into the formula for the area, we obtain an alternative formula for the area of a semi-circle using the diameter:

Area=$\frac{1}{2}\:\pi\:(\frac{d}{2})^2$

**What is the Area Formula of a Semi-Circle?**

Through the derivation we have previously done, the area A of a semi-circle whose radius r is known can be expressed using the formula:

Area=½ πr^{2}

If the radius is not known but a diameter d is provided, we have an alternative formula that is expressed in terms of the diameter:

Area=$\frac{1}{2}\:\pi\:(\frac{d}{2})^2$

We note that for both formulas, the **unit of the area **is expressed in terms of **squared units** of the given radius/diameter.

**How is it Used?**

Again, we practice what we have learned so far by working on the same guided example as earlier, but for finding the area of the semi-circle:

From the given figure, we are given the radius of the semi-circle to be three units. Hence, we can say that:

r=3 units

Moreover, since the radius is known, we can use the Area Formula in terms of the radius r:

Area=½ πr^{2}

We then substitute the given radius to the formula, along with an approximate value of π≈3.14 to get:

A=½×3.14×(3 units)^{2}

Afterward, we take the square of the radius to get:

A=½×3.14×9 units^{2}

Then, multiplying the numbers together we obtain the area of the semi-circle:

A=14.13 units^{2}

We then conclude that the area of the semi-circle is 14.13 units^{2}. This quantity is expressed in terms of squared units of the radius.

## Problem-Solving Examples

We can now proceed to solve sample problems to apply what we have learned so far. Each problem tackles different formulas discussed and gives us a challenge on how to solve through the information given to us.

**Perimeter of a Semi-Circle**

**Sample Problem 1:**

What is the perimeter of a semicircular disc whose radius is four centimeters?

**Solution:**

We recall that the perimeter P of a semicircle is given by the formula:

P=r×(π+2)

With the given radius r=4 cm and an approximate value of π≈3.14, we substitute these values into the formula:

P=4 cm×(3.14+2)

Adding the numbers inside the parentheses, we have:

P=4 cm×(5.14)

Finally, by multiplying the radius we obtain the value of the perimeter to be:

P=20.56 cm

Therefore, we conclude that the perimeter of the semicircular disc is 20.56 cm.

**Sample Problem 2:**

A painter wants to measure the dimensions of a semicircular glass pane using only a meterstick. If he measures the straight edge to be 15 inches, what would be the perimeter of the glass pane?

**Solution:**

We observe that the straight edge of the glass pane measures the diameter of the semicircle. With a diameter given instead of a radius, we can use the other formula for the perimeter P of a semicircle:

P=$\frac{d}{2}$×(π+2)

Using the above formula, we substitute the given diameter d=15 in. We also note that an approximate value of π=3.14 is used:

P=$\frac{15 in}{2}$×(3.14+2)

Then, we add the numbers enclosed in parentheses:

P=$\frac{15 in}{2}$×(5.14)

We proceed by dividing the given diameter by two:

_{P=7.5 in×(5.14)}

Lastly, we multiply the two numbers to get the value of the perimeter:

P=38.55 in

Therefore, the perimeter of the glass pane is 38.55 inches.

**Sample Problem 3:**

Teacher Lily is designing a creative activity for the students in her math class. She wants everyone to decorate measuring tools, including a semicircular protractor. If she asks the class to wrap a string around the edges of the tools, and the protractor has a radius of 2 inches, how much string does each student need to decorate the protractor?

**Solution:**

We note that the radius of the semicircular protractor is given by r=2 in. For a given radius, the perimeter P can be computed using the formula:

P=r×(π+2)

Hence, we can substitute the given values, with π≈3.14 into the above equation to get:

P=2 in×(3.14+2)

We then simplify the equation by taking the sum of the numbers inside the grouping symbols:

P=2 in×(5.14)

Lastly, we multiply the given radius by the sum obtained earlier to determine the value of the perimeter of the protractor:

P=10.28 in

Since the perimeter of the protractor is 10.28 inches, then the amount of string required for each student to decorate it is also 10.28 inches.

**Area of a Semi-Circle**

**Sample Problem 4:**

Suppose we use the same protractor from Sample Problem 3 whose radius is given by 2 inches. What is the area of the protractor?

**Solution:**

We recall that the area A of a semicircle is given by the formula:

A=½πr^{2}

With the given radius r=2 in, and an approximate value of π≈3.14, we substitute these into the formula to get:

A=½×3.14×(2 in)^{2}

Getting the square of the radius, we have:

A=½×3.14×4 in^{2}

Finally, by multiplying the three numbers we can compute for the value of the area:

A=6.28 in^{2}

Hence, the area of the protractor is 6.28 in^{2}.

**Sample Problem 5:**

Clark ordered a round 12″ pepperoni pizza and decided to share half of it with his neighbor Julie. He wants to know how much area the half-pizza takes up so he can tell it in advance to Julie. What is the area of the 12’’ pizza?

**Solution:**

We first observe that the diameter of the pepperoni pizza is given to us. For a given diameter d, the area A of the semicircle can be expressed using the formula:

$\frac{1}{2}\:\pi\:(\frac{d}{2})^2$

Substituting the given diameter d=12 in, and using an approximate value of π≈3.14, we plug these values in the formula to get:

A=½×3.14×$({12 in}{2})^2$

Dividing the diameter by two, we simplify the fraction inside the parentheses. We note that we are effectively taking the radius of the semi-circle using the given diameter:

A=½×3.14×(6 in)^{2}

Then, we proceed by taking the square of the radius:

A=½×3.14×36 in^{2}

Thus, the value of the area can be obtained by multiplying the numbers:

A=56.52 in^{2}

Therefore, the area of the pepperoni pizza is 56.52 in^{2}.

**Sample Problem 6:**

Suppose a baker wants to bake round cookies, to be split equally into half-cookies. Each cookie should measure 3inches in radius, and ten grams of flour is required for each square inch of cookie to be baked. How much flour does the baker need to make a dozen half-cookies?

**Solution:**

We first note that the baker needs to make a dozen semicircular cookies of radius r=3 in. We are asked to determine the amount of flour required to bake this many cookies, based on the total area to be baked. Hence, we first solve for the total area covered by the cookies, then apply ratio and proportion to solve for the amount of flour needed by the baker.

To get the area of each half-cookie, we use the formula for the area A of a semi-circle given its radius r:

A=½ πr^{2}

We then substitute the known radius, and use an approximate value of π≈3.14 into the formula:

A=½×3.14×(3 in)^{2}

Taking the square of the radius, we have:

A=½×3.14×9 in^{2}

Hence, the area of each half-cookie can be obtained by multiplying together these numbers:

A=14.13 in^{2}

Since each half-cookie has an area of 14.13 in^{2}, the total area AT of a dozen cookies can be computed by multiplying the area of one half-cookie by one dozen:

A_{T}=A×12

A_{T}=14.13 in^{2}×12=169.56 in^{2}

To determine how much flour the baker needs, we apply the flour-to-cookie ratio of 10 grams :1 in^{2}, as shown below:

10 grams flour=1 in^{2} cookie

For the total area A_{T}=169.56 in^{2}, the amount of flour F required is given by:

F=169.56 in^{2} x ($\frac{10 g}{1 in^2}$)=1695.6 g

Therefore, the baker needs 1659.6 grams, or 1.6596 kilograms** **of flour to bake a dozen half-cookies.

## Summary

A **semi-circle **is **half of a circle**. It is formed by an **arc** covering half of a circle, and a **diameter** connecting both ends of the arc.

Like circles, any point along the arc of a semi-circle is **equidistant **from its center. A line from the center to any point in the arc is a **radius** of the semi-circle.

A **perimeter** is a **measure of the boundary enclosing a shape**. It can be obtained by taking the length of the edges enclosing the shape.

On the other hand, an **area** is a **measure of the space a shape occupies in a 2D plane**. This quantity is dependent on the dimensions of the shape.

The mathematical constant **pi** can be expressed as the ratio between the circumference of a circle and its diameter. It can be approximated as π≈3.14.

The **perimeter** P of a semi-circle with a known radius r is given by the following formula:

P=r×(π+2)

Alternatively, its perimeter can also be expressed in terms of the diameter d:

P=$\frac{d}{2}$ x (π+2)

The **area **A of a semi-circle whose radius r is known can be determined using the formula:

A=½πr^{2}

Alternatively, its area can also be expressed in terms of the diameter d:

A=$\frac{1}{2}\:\pi\:(\frac{d}{2})^2$

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