Area of Ring (Annulus): Formula, Derivation and Solved Examples

The area of ring, also known as an annulus, is the space between two concentric circles with different radii. It is a key concept in geometry that appears in many real-world applications, such as designing circular tracks, washers, and rings. To find the area of an annulus, you subtract the area of the smaller inner circle from the larger outer circle. With this approach, you can quickly determine the area of the ring without needing complex methods, making it an essential concept in basic and advanced mathematics. In this guide, you will learn how to calculate the area of an annulus using a simple formula, along with clear explanations and examples for better understanding.

Table of Contents

• What is a Ring (Annulus)?
• Area of Ring: Formula
• Derivation of the Area of Ring Formula
• Real-Life Examples of an Annulus
• Solved Examples on Area of Ring
• Practice Questions on the Area of Ring

What is a Ring (Annulus)?

A ring (also called an annulus) is the region between two concentric circles. The word 'annulus' comes from the Latin word meaning ‘little ring'. The area of a ring is the difference between the areas of the outer and inner circles. These two circles are called 'concentric circles'; 'concentric' simply means they share a common centre.

In the diagram above, both circles share the same centre point O. The larger circle has radius R (called the outer radius), and the smaller circle has radius r (called the inner radius). The shaded region between them is the annulus, or the area of the ring. R > r. The outer radius is always larger than the inner radius.

Area of Ring: Formula

The area of a ring (annulus) is the space enclosed within its boundaries, essentially, subtracting the area of the inner circle (the hole) from the area of the outer circle. The formula is:

Area of Ring (Annulus), A = π(R² − r²)

Derivation of the Area of Ring Formula

Let the outer radius be R and the inner radius be r. The area of a ring (annulus) is found by subtracting the area of the smaller circle from the larger circle. Here are the steps to derive the formula:

Step 1: Area of the outer circle (radius R) = πR²

Step 2: Area of the inner circle (radius r) = πr²

Step 3: Area of the annulus = Area of outer circle − Area of inner circle

= πR² − πr²

Step 4: Take π common: = π(R² − r²)

Step 5: Apply the identity a²−b² = (a+b)(a−b):

π(R² − r²) = π(R + r)(R − r)

Real-Life Examples of an Annulus

The annulus (ring) shape is everywhere around us. Here are a few real-life examples of ring shapes:

• Metal Washer: The face of a bolt washer is a perfect annulus. Engineers use the area formula to calculate the material needed.

• CD / DVD: The readable surface of a compact disc is an annulus: a circle with a hole in the middle.

• Doughnut (top view): Seen from above, a doughnut is an annulus. The area of the icing on top can be calculated using the annulus formula.

• Saturn's Rings: Saturn's ring system forms a giant annulus in space. Astronomers use this concept to calculate the area of the rings.

• Pipe Cross-section: When you cut a hollow pipe, the cross-section is an annulus. Plumbers use this to find the flow area of pipes.

• Watch Face / Ring: A finger ring or the bezel of a watch forms an annular shape. Jewellers calculate material weight from its area.

Solved Examples on Area of Ring

Example 1: A circular path runs around a round garden. The outer radius of the path is 21 m and the inner radius (the garden's radius) is 14 m. Find the area of the circular path. (Use π = 22/7)

Solution: Given: Outer radius R = 21 m, Inner radius r = 14 m

Area = π(R² − r²)

Substituting the values: Area = (22/7) × (21² − 14²) = (22/7) × (441 − 196) = (22/7) × 245 = 22 × 35 = 770 m²

Area of the circular path = 770 m²

Example 2: A hollow steel pipe has an outer diameter of 10 cm and an inner diameter of 6 cm. Find the area of the cross-section. (Use π = 3.14)

Solution: Given D = 10 cm and d = 6 cm.

Therefore, R = 10 ÷ 2 = 5 cm  and  r = 6 ÷ 2 = 3 cm

Area = π(R² − r²) = 3.14 × (5² − 3²) = 3.14 × (25 − 9) = 3.14 × 16 = 50.24 cm²

Cross-sectional area of the pipe = 50.24 cm²

Example 3: The area of an annulus is 770 cm², and its outer radius is 21 cm. Find its inner radius. (Use π = 22/7)

Solution: Given: Area = 770 cm², R = 21 cm.

We need to find r.

Area = π(R² − r²)

770 = (22/7) × (R² − r²) = (22/7) × (441 − r²)

441 − r² = 770 × 7/22 = 5390/22 = 245   

r² = 441 − 245 = 196

r = √196 = 14 cm

Example 4: A circular swimming pool of radius 7 m is surrounded by a 3.5 m wide footpath. Find the cost of paving the footpath at ₹200 per m². (Use π = 22/7)

Solution: Given the inner radius r = 7 m (the pool). 

Outer radius R = 7 + 3.5 = 10.5 m (pool + path)

Area of footpath: A = π(R² − r²) = (22/7) × (10.5² − 7²) = (22/7) × (110.25 − 49) = (22/7) × 61.25

= 22 × 8.75 = 192.5 m²

The cost of paving the footpath = ₹200 per m².

Cost of paving 192.5 m² footpath = 192.5 × ₹200 = ₹38,500

The total cost of paving the footpath is ₹38,500.

Example 5: Find the area of the path, where a path is 14 cm wide and surrounds a circular lawn whose diameter is 360 cm.

Solution: Given: Width of path = 14 cm

Diameter of lawn = 360 cm ⇒ Inner radius (r) = 180 cm

Outer radius (R) = 180 + 14 = 194 cm

Area of path = π (R² − r²)

= π (194² − 180²)

= π (37636 − 32400)

= π (5236)

= 5236π cm²

Taking π = 22/7

Area of path = 5236 × 22/7

= 16459.43 cm² (approx.)

Therefore, Area of the path = 5236π cm² ≈ 16459.43 cm²

Practice Questions on the Area of a Ring

1. A bangle has an outer radius of 5 cm and an inner radius of 4.5 cm. Find the area of the metal used to make the bangle. (Use π = 3.14) 

2. A circular pond of diameter 28 m is surrounded by a 3.5 m wide path. Find the area of the path. (Use π = 22/7)

3. The inner and outer radii of a circular track are 56 m and 63 m, respectively. Find (i) the area of the track and (ii) the cost of levelling it at ₹5 per m². (Use π = 22/7)

4. A circular ring has an outer radius of 12 cm and an inner radius of 9 cm. Find the area of the metal used to make the ring. (Use π = 3.14)

5. The outer radius of a circular ring is 25 cm and its inner radius is 18 cm. Find: (i) Area of the ring (ii) If the cost of polishing is ₹2 per cm², find the total cost. (Use π = 3.14)

6. A circular lake has a radius of 50 m. A 10 m wide strip of land surrounds it. Find the area of the strip. (Use π = 22/7)

7. A circular field of radius 30 m has a 4 m wide road around it. Find the area of the road. (Use π = 22/7)

8. A circular garden of diameter 40 m is surrounded by a 5 m wide walking track. Find the area of the path. (Use π = 22/7)

9.  A circular path has an outer radius of 20 cm and its area is 314 cm². Find the inner radius. (Use π = 3.14)

10. The area of a ring is 616 cm². If the outer radius is 15 cm and π = 22/7, find the inner radius.