Area of a Circular Ring (Annulus): Class 10 Maths

Quick answer: The region between two concentric circles is a circular ring (annulus). Its area is the outer circle's area minus the inner circle's: $A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \pi(R - r)(R + r)$, where $R$ is the outer radius and $r$ the inner radius. For a park of radius 21 m with a 7 m wide track around it, $R = 28$ m and the track's area is $\frac{22}{7} \times 7 \times 49 = 1078$ m².

Syllabus note: In the rationalised NCERT textbook, the section "Areas of combinations of plane figures" was removed from Areas Related to Circles; areas of sectors and segments remain. Ring problems use only circle areas, but check whether your board still sets them.

Parts of a ring

Area of a ring

Step 1: Area of the outer circle $= \pi R^2$.

Step 2: Area of the inner circle $= \pi r^2$.

Step 3: Subtract:

$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$

Step 4: Factorise with $a^2 - b^2 = (a - b)(a + b)$:

$A = \pi(R - r)(R + r)$

The factorised form saves work: with $\pi = \frac{22}{7}$, a width of 7 cancels the 7 in the denominator straight away.

Circumferences

The outer edge is longer than the inner edge by $2\pi w$, whatever the size of the circles.

Worked example: park and running track

A circular park of radius 21 m is surrounded by a running track 7 m wide. Find (i) the area of the track, (ii) the cost of paving it at ₹50 per m², and (iii) the cost of fencing the outer edge at ₹20 per m. Take $\pi = \frac{22}{7}$.

Step 1: $r = 21$ m and $w = 7$ m, so $R = 21 + 7 = 28$ m.

Step 2: Area of the track:

$A = \frac{22}{7} \times (28 - 21) \times (28 + 21) = \frac{22}{7} \times 7 \times 49 = 1078 \text{ m}^2$

Step 3: Cost of paving $= 1078 \times 50 =$ ₹53,900.

Step 4: Outer circumference:

$2\pi R = 2 \times \frac{22}{7} \times 28 = 176 \text{ m}$

Step 5: Cost of fencing $= 176 \times 20 =$ ₹3,520.

Answer: (i) 1078 m², (ii) ₹53,900, (iii) ₹3,520.

Check: $28^2 - 21^2 = 784 - 441 = 343$ and $\frac{22}{7} \times 343 = 1078$.

Summary table

Exam tips

• If a diameter is given, halve it to get the radius before adding the width.
• Use the factorised form to avoid squaring large numbers.
• Check whether the question asks for the inner edge, the outer edge or both when finding fencing costs.

Common mistakes

• Adding the width to the diameter instead of the radius.
• Writing $\pi(R - r)^2$ instead of $\pi(R^2 - r^2)$.
• Using the inner circumference for fencing the outer edge.
• Mixing units, such as a radius in metres and a rate per cm².

Frequently Asked Questions

What is an annulus?

It is the ring-shaped region between two concentric circles.

What is the formula for the area of a circular ring?

Area = π(R² − r²) = π(R − r)(R + r), where R is the outer radius and r the inner radius.

How do you find the outer radius of a track around a park?

Add the width of the track to the radius of the park: R = r + w.

What is the area of a 7 m wide track around a park of radius 21 m?

The outer radius is 28 m, so the area is (22/7) × 7 × 49 = 1078 m².

Is the difference between the outer and inner circumference always the same for a given width?

Yes. It is 2πw, which depends only on the width of the ring.