JUMP TO TOPIC

**Circumradius|Definition & Meaning**

**Definition**

In **polygons**, the **circumradius equals** the **radius** of the **circle** that **touches the vertices** of the **polygon** or the **circle** in which the polygon can be inscribed. **When** a **triangle** is **circumscribed** by a circle, then the **circumradius **is the** radius** of that **circle**. Due to the **cyclical nature of triangles**, **each triangle** has a **circumradius** but this is not the case with polygons.

**Building Concept for Circumradius**

A** glass ceiling** is shaped like a **pentagon** with equal sides, and an **architect designs** it so that it is **dome-shaped** and has a **beam of a circle** around it. As depicted in the figure, the **designer realizes** that** each** of the **pentagonal windows** will **need** to **have supportive beams** going **from** its **vertex** to the **mid**(center) of the beam of the circle when designing the glass ceiling.

**Circumcircle**

Mathematics is actually a fascinating subject when we talk about this **pentagonal glass ceiling** and its **circular beam.** As a result of passing through** each vertice** of a polygon, a **circumcircle** is a circle **drawn** around it. The **circular beam** **constitutes** the **circumcircle** in this example, while the **Polygon** is **represented** by a **pentagonal ceiling.**

**Circumradius**

**Polygons** with **circumcircles** also possess a **circumradius**, which is another important characteristic. An arbitrary polygon’s **circumradius** is the **radius of the circle** inside **which** the **polygon** is **inscribed**. In other words, it is a **line segment** that **connects** any **vertex of the polygon** to the **circumcircle’s center.**

There is a** circumradius** on **each triangle** and **tetrahedron,** but **not for all polygons or polyhedra**. The **few polygons** that **have** a **circumradius** are **regular polygons**.

**Circumradius vs Inradius**

The **circumradius and inradius** of a triangle are different types of radii. When a triangle is **circumscribed by a circle**, then the **circumradius** is the **radius of that circle**. In a triangle, the** circumcenter** is where all three vertices of the triangle touch, and these points are called the triangle’s vertices. Alternatively, an **inradius** indicates the **distance between** the **triangle-enclosed circle’s center** and the **tangent point** of that triangle.

**Circumradius of Triangle**

From the **vertices of the triangles**, we calculate the **circumradius**, which **is equal** to the **distance between** **them** **and** the **circle’s center**.** Circumcircle** is another name for this circle. This is shown in the figure below.

**Euler Theorem**

According to **Euler’s triangle theorem**, the **circumradius** **and inradius** of a triangle **are related** to the distance between their centers.

Let **r** be the **inradius** (intersection point of all three vertices of the triangle).

Let **R** be the **circumradius** (When a triangle is circumscribed by a circle, then the circumradius is the radius of that circle).

Let the **distance** between the circumcenter**(R)** **and **incenter**(r)** be **d**. We can write

**$d^{2}=R(R-2r)$**

**The Formula of the Circumradius of the Triangle**

Suppose the sides of the triangle as a**, b, and c** and we want to find the **circumradius** so** we can write** it as:

**Circumradius of Triangle** =$\dfrac{abc}{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}} $

**Finding Circumradius of Triangle Practically**

There are relatively** few steps** involved in the** construction** of a **circumradius**. **Compass** and** straight edge** are the only materials you’ll need.

**Step 1**

Draw a **triangle** with three vertices** L, M, and N.**

**Step 2**

The next step is to** draw a circle** that **touches** every **vertex** or side of the **triangle**. The **circumcircle** can be** seen** here. This circle has a circumcenter in its center.

**Step 3**

From **one of the three** **vertices,** **draw** a **segment** around the circumcenter. Measurement of the** circumradius** is based on the length of this line segment.

**Step 4**

Join **every vertex** of the **triangle** with** point O** so that we get **OA, OB, and OC**. OA, OB, and OC are known as **circumradii** of the triangle shown in the figure.

**Circumradius of Regular Polygon**

In euclidean geometry, a **regular polygon** is one that is **direct equiangular.** This means that all angles are equal in measure, and **equilateral** means all sides have the same length.

**Consider** the **following polygon** having **each vertex **of the **same length** and a **circle is present inside** which the polygon is inscribed. The **line** segment** from** the** center** of that** circle** **meeting** on any end of the **polygon vertex** is known as a **circumradius** as shown in the figure.

**The Formula of Circumradius of a Regular Polygon**

Let **a** be **each vertex o**f the polygon being equal in length and **n** is the **number of the vertex** in the polygon then the **circumradius of a polygon** can be **written** as:

**Circumradius of Regular Polygon** =$\dfrac{a}{2\sin\frac{\pi}{n}}$

**Circumradius of Cyclic Quadrilateral**

**Cyclic quadrilaterals** can be **scribed** into **circles** by inlaying **four sides.** Q**uadrilaterals** have **four chords** **connecting** their **vertex** to the **circle circumference**. The** radius** of that **circle** is **known** as the **circumradius** of that cyclic quadrilateral. Below is an illustration of the concept.

**The Formula of Circumradius of Cyclic Quadrilateral**

Consider a **cyclic quadrilateral** having four vertex **AB, BC, CD, and DA**, and suppose that

**a = AB, b = BC , c = CD and d = DA**

We can write the formula for the **semi-perimeter** (which will later be used in circumradius)

**Semiperimeter of cyclic quadrilateral **= $\displaystyle\frac{a+b+c+d}{2} $

From this, we can write the formula for cyclic quadrilateral as

**Circumradius of cyclic quadrilateral** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}}$

**Some Examples of Finding the Circumradius**

**Example 1**

Consider a **Triangle** with sides **a=5cm, b=6cm, c=7cm**, and a circle is circumscribed. Find** the circumradius.**

**Solution**

**Circumradius R** = $\dfrac{abc}{\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}} $

**Circumradius R** = $\dfrac{5 \cdot 6 \cdot 7}{\sqrt{(5+6+7)(6+7-5)(7+6-5)(5+6-7)}}$

**Circumradius R** = $\dfrac{210}{\sqrt{(18)(6+7-5)(7+6-5)(5+6-7)}} $

**Circumradius R** = $\dfrac{210}{\sqrt{(18)(8)(8)(4)}}$

**Circumradius R** = $\dfrac{210}{\sqrt{(4608)}} $

**Example 2**

Consider **six sides** of **regular polygon** **AB, BC, CD, DE, EF, and FG,** which are equal in length. Find** the circumradius**.

**Solution**

We are given the data:

**a = **5 cm

**n = **6

**Circumradius R** = $\dfrac{a}{2\sin\frac{\pi}{n}} $

**Circumradius R** = $\dfrac{5}{2\sin\frac{\pi}{6}} $

**Circumradius R** = $\dfrac{5}{2 \times 0.00913} $

**Circumradius R** = $\dfrac{5}{0.0182} $

**Example 3**

Consider a **cyclic quadrilateral** with **a=3, b=4, c=5, and d=6** **find the circumradius**.

**Solution**

**First** we will find **semiperimeter:**

**Semiperimeter** = $\dfrac{3+4+5+6}{2} $

**Semiperimeter** = $\dfrac{18}{2} $

**Semiperimeter** = $9$

**Circumradius R** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{(ab+cd)(ac+bd)(ad+bc)}{(s-a)(s-b)(s-c)(s-d)}} $

**Circumradius R** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{(3 \times 4+5 \times 6)(3\times 5+4\times 6)(3\times 6+4\times 5)}{(9-3)(9-4)(9-5)(9-6)}} $

**Circumradius R** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{(12+30)(15+24)(18+20)}{(6)(5)(4)(3)}} $

**Circumradius R** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{(42)(39)(38)}{360}} $

**Circumradius R** = $\displaystyle\frac{1}{4} \times \sqrt{\frac{62244}{360}} $

**Circumradius R** = $\displaystyle\frac{1}{4} \times 13.149 $

**Circumradius R** = $3.28$

*All mathematical drawings and images were created with GeoGebra.*