**Introduction**

A semicircle is a fundamental geometric concept students learn to understand and apply in various mathematical scenarios. This in-depth article seeks to give readers an overview of the topic and its main ideas and practical applications. We will explore the topic’s grade appropriateness, related math domain, Common Core Standards, and practical examples to help students and educators.

**Grade Appropriateness**

The concept of a semicircle is typically introduced in middle school, around the 6th or 7th grade, and extends into high school geometry courses.

**Math Domain**

The semicircle belongs to the domain of geometry, a branch of math that deals with shapes, sizes, and properties of figures.

**Applicable Common Core Standards**

The relevant Common Core Standards for studying semicircles are:

*7.G.B.4:* Know the formulas for the area and circumference of a circle and use them to solve problems.

*G.C.A.2:* Identify and describe relationships among inscribed angles, radii, and chords.

**Definition **

A ** semicircle** is a shape formed by cutting a circle along its diameter, resulting in half of the circle. It is a two-dimensional figure consisting of a curved line (the arc) and a straight line (the diameter).

**Key Concepts**

**Diameter***:* The diameter is a line that passes through the center of the circle and connects two points on the circle.

** Radius:** Any segment that connects the circle’s center to any other point on the circle is known as the radius.

** Arc:** The arc is the curved portion of the circle’s circumference that forms the boundary of the semicircle. If the circle has an arc measuring 360°, a semicircle has an arc that measures 180°.

*Properties:*

A semicircle is always a closed 2D shape.

Due to its curved edge, a semicircle can never be a polygon.

A semicircle has only one curved edge, its arc, and a straight edge, its diameter.

It is always exactly half of a circle.

Half of a circle’s area is the area of the semicircle.

**Discussion with Illustrative Examples**

**Example 1:** *The relationship between the radius and diameter of a semicircle.*

In a semicircle, the diameter is twice the radius’s length, connecting two points on the circle and going through the center.

**Example 2:** *Calculating the length of the arc.*

The length of the arc of a semicircle is half the circumference of the full circle. C=2πr is the formula for calculating the circumference of a circle, so the length of the arc of a semicircle is ½(2πr)=πr.

**Example 3: ***Calculating the area and perimeter*

How did we get the formulas, you ask? Since we know that the formula for calculating the area of a circle is obtained by getting the product of and the square of its radius; therefore, to get the area of a semicircle, we divide it by 2. Meanwhile, suppose the circumference for a circle is two multiplied by and the radius when we divide it by two and add the straight edge diameter. In that case, we get pi multiplied by radius plus twice the radius!

**Examples with Solution**

**Example 1**

A semicircle has a diameter of 20 cm; find its radius and arc length.

**Solution**

Radius=$\frac{diameter}{2}$

Radius=$\frac{20}{2}$

Radius=10 cm

Arc length of the semicircle=πr

Arc length of the semicircle=(π) (10)

Arc length of the semicircle≈31.42 cm

Therefore, a semicircle with a diameter of 20 cm has a radius of **10 cm** and an arc length of approximately **31.42 cm**.

**Example 2**

Determine the area and perimeter of a semicircle with a radius of 4 cm.

**Solution**

Area=½πr^{2}

Area=½⋅π⋅4^{2}

Area≈25.13 cm^{2}

Perimeter=πr+2r

Perimeter=(π)(4)+(2)(4)

Perimeter≈20.57 cm

Hence, the area and perimeter of a semicircle with a radius of 4 cm are **25.13 cm**** ^{2}** and

**20.57 cm**, respectively.

Now that we know the semicircle formulas, how about we apply them?

**Real-life Application with Solution**

**Example 1**

We have a semicircular table. If the radius of the table is 2 ft, what are its area and perimeter? *Use 3.14 as the approximate value of π.*

**Solution**

**Example 2**

A semicircular garden has a diameter of 10 meters. Calculate the area of the garden and the length of the fence needed to surround it.

**Solution**

Radius=$\frac{Diamter}{2}$=$\frac{10}{2}$=5 meters

Area=½πr^{2}

Area=½(π)(5^{2})

Area=39.27 m^{2}

Perimeter=πr+2r

Perimeter=(π)(5)+(2)(5)

Perimeter=25.71 m

The garden has an area of approximately **39.27 square meters**, and a fence of approximately **25.71 meters** is needed to surround it.

**Practice Test**

1. Find the semicircle’s perimeter with a diameter of 12 meters.

2. Calculate the area of a semicircle with a radius of 7 cm.

3. Determine the length of the arc of a semicircle with a radius of 8 cm.

4. Calculate the area of a semicircle with a circumference of 18.85 cm.

5. A semicircular window has a diameter of 8 feet. What is the area of the window?

**Frequently Asked Questions (FAQs)**

### What is the difference between a semicircle and a half-circle?

A semicircle and a half-circle refer to the same shape, half of a circle. The terms are used interchangeably.

### How do you find the area and perimeter of a semicircle?

To find the area of a semicircle, use the formula Area=½πr^{2}, where r is the radius.

To find the perimeter, use the formula Perimeter = πr+2 r, where r is the radius.

### Can a semicircle be inscribed in a triangle?

Yes, a semicircle can be inscribed in a triangle, known as an inscribed semicircle. The diameter of the inscribed semicircle will be equal to the length of the base of the triangle, and the vertices of the triangle will be tangent to the semicircle.

### Can a semicircle be a sector?

A semicircle is a specific type of sector where the central angle is 180 degrees. A sector is a portion bounded by two radii and an arc.

### Are the area and perimeter formulas for a semicircle applicable to other shapes?

The area and perimeter formulas for a semicircle are specific to semicircles. However, the concepts and techniques used to derive these formulas can be applied to other shapes, such as sectors and segments of circles, by adapting the formulas accordingly.

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