Areas Related to Circles • Topic 1 of 3

Perimeter and area of a circle

What is the perimeter and area of a circle? The perimeter of a circle is the total distance around its outer boundary. In geometry, this specific boundary length is called the circumference. The area of a circle represents the total amount of flat space enclosed inside that boundary. Imagine a circular running track: if you run all the way around the outer white line, you have covered the circumference. If you need to cover the grass field inside the track with fresh turf, you are calculating the area.

To measure these values, mathematicians use a special constant called Pi (written as the Greek symbol $\pi$). Pi represents a fixed ratio: the circumference of any circle divided by its diameter. No matter how small a coin or how massive a ferris wheel is, this ratio is always the same! For calculations, we approximate $\pi$ as 22/7 or 3.14.

  • Radius (r): The straight-line distance from the exact center of the circle to any point on its outer edge.
  • Diameter (d): The maximum straight distance across a circle, passing through the center. It is always equal to twice the radius ($d = 2r$).

Formulas for calculations:

  • Circumference of a circle = $2 \cdot \pi \cdot r$
  • Area of a circle = $\pi \cdot r^2$
MeasurementPhysical MeaningFormulaPrimary Units
CircumferenceOuter boundary line length$2 \cdot \pi \cdot r$cm, m, km (linear units)
AreaInside flat space surface$\pi \cdot r^2$sq. cm, sq. m (square units)

Deriving area from circumference

Why is the area $\pi r^2$? Slice a circle into many thin sectors and rearrange them alternately to form a near-rectangle. The rectangle's length is half the circumference, $\frac{1}{2}(2\pi r)=\pi r$, and its breadth is the radius $r$. Area $=\text{length}\times\text{breadth}=\pi r\times r=\pi r^2$. The thinner the slices, the closer the shape gets to a true rectangle, which is why the formula is exact.

Distance covered by a rolling wheel

When a wheel makes one complete turn, every point on its rim returns to its start, and the wheel advances forward by exactly one circumference. So:

$$\text{Distance} = (\text{Number of revolutions}) \times (\text{Circumference})$$

Rearranged, $\text{Revolutions} = \dfrac{\text{Total distance}}{2\pi r}$. Always convert speed (km/h) to distance first, and keep every length in the same unit before dividing.

Area of a ring (annulus)

A circular path, washer, or pipe cross-section is the region between two concentric circles. If the outer radius is $R$ and the inner radius is $r$, the shaded ring area is the big circle minus the small one:

$$\text{Ring area}=\pi R^2-\pi r^2=\pi\left(R^2-r^2\right)=\pi(R+r)(R-r).$$

For a path of uniform width $w$ around a circle of radius $r$, the outer radius is $R=r+w$.

Two circles combined into one

If two circles of radii $r_1$ and $r_2$ are melted/recast into a single circle of radius $R$, the property that is conserved decides the equation. If their areas add up: $\pi R^2=\pi r_1^2+\pi r_2^2\Rightarrow R^2=r_1^2+r_2^2$. If their circumferences add up: $2\pi R=2\pi r_1+2\pi r_2\Rightarrow R=r_1+r_2$.

Common mistakes to avoid

  • Radius vs diameter: a question often gives the diameter. Halve it first. A 70 cm wheel has $r=35$ cm, not 70.
  • Choosing $\pi$: use $\pi=\frac{22}{7}$ when the radius is a multiple of 7 (the 7 cancels neatly); use $\pi=3.14$ otherwise, or when the problem says so. Never mix both in one calculation.
  • Units: circumference is a length (cm, m), area is square units ($\text{cm}^2$, $\text{m}^2$). Convert km to m and m to cm before computing revolutions.
  • Scaling: doubling $r$ doubles the circumference but multiplies the area by $4$, because area depends on $r^2$.
A circle of radius r with its circumference C = 2πr as the boundary and area A = πr squared shaded insideBoundary length C = 2πr · inside area A = πr^2rA = πr^2centre OC = 2πr (boundary)rrdiameter d = 2rA ring (annulus): the shaded region between an outer circle of radius R and an inner circle of radius rRing (annulus) area = π(R^2 − r^2)Rrouter circle − inner circle= π(R + r)(R − r)
1
Worked Example
A circular garden has a radius of 14 meters. Find the total distance a gardener walks to complete one full lap around its outer fence, and calculate the total surface area of the grass inside. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Given radius $r=14$ m.
  2. Step 2: Circumference $=2\pi r=2\times\frac{22}{7}\times 14=2\times 22\times 2=88$ m.
  3. Step 3: Area $=\pi r^2=\frac{22}{7}\times 14\times 14=22\times 2\times 14=616\ \text{m}^2$.

Answer: Circumference $=88$ m and area $=616\ \text{m}^2$.

2
Worked Example
The outer boundary length of a circular pond is 176 m. Find the radius of the pond and its area. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: $2\pi r=176\Rightarrow 2\times\frac{22}{7}\times r=176$.
  2. Step 2: $\frac{44}{7}r=176\Rightarrow r=176\times\frac{7}{44}=4\times 7=28$ m.
  3. Step 3: Area $=\pi r^2=\frac{22}{7}\times 28\times 28=22\times 4\times 28=2464\ \text{m}^2$.

Answer: Radius $=28$ m and area $=2464\ \text{m}^2$.

3
Worked Example
A bicycle wheel has diameter 70 cm. How many complete revolutions must it make to cover 1.1 km? (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Diameter $=70$ cm, so radius $r=35$ cm.
  2. Step 2: Convert the distance: $1.1$ km $=1100$ m $=110000$ cm.
  3. Step 3: Distance in one revolution $=$ circumference $=2\pi r=2\times\frac{22}{7}\times 35=220$ cm.
  4. Step 4: Revolutions $=\frac{110000}{220}=500$.

Answer: $500$ revolutions.

4
Worked Example
Find the area of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 19 cm and 9 cm. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: When circumferences add, $2\pi R=2\pi r_1+2\pi r_2$, so $R=r_1+r_2$.
  2. Step 2: $R=19+9=28$ cm.
  3. Step 3: Area $=\pi R^2=\frac{22}{7}\times 28\times 28=22\times 4\times 28=2464\ \text{cm}^2$.

Answer: $2464\ \text{cm}^2$ (the new circle has radius $28$ cm).

5
Worked Example
The radii of two circles are 8 cm and 6 cm. Find the radius of the circle whose area equals the sum of the areas of these two circles.
Solution
  1. Step 1: When areas add, $\pi R^2=\pi r_1^2+\pi r_2^2$, so $R^2=r_1^2+r_2^2$.
  2. Step 2: $R^2=8^2+6^2=64+36=100$.
  3. Step 3: $R=\sqrt{100}=10$ cm.

Answer: Radius $=10$ cm.

6
Worked Example
A circular park has radius 21 m. A 3.5 m wide gravel path runs all the way around it. Find the area of the path. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Inner radius $r=21$ m; outer radius $R=21+3.5=24.5$ m.
  2. Step 2: Path area $=\pi(R^2-r^2)=\pi(R+r)(R-r)$.
  3. Step 3: $=\frac{22}{7}\times(24.5+21)\times(24.5-21)=\frac{22}{7}\times 45.5\times 3.5$.
  4. Step 4: $=\frac{22}{7}\times 159.25=22\times 22.75=500.5\ \text{m}^2$.

Answer: $500.5\ \text{m}^2$.

7
Worked Example
The area of a circular plate is $154\ \text{cm}^2$. Find its circumference. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: $\pi r^2=154\Rightarrow\frac{22}{7}r^2=154$.
  2. Step 2: $r^2=154\times\frac{7}{22}=7\times 7=49\Rightarrow r=7$ cm.
  3. Step 3: Circumference $=2\pi r=2\times\frac{22}{7}\times 7=44$ cm.

Answer: Circumference $=44$ cm.

8
Worked Example
A wheel of a car makes 5000 revolutions to cover a distance of 11 km. Find the radius of the wheel. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Distance $=11$ km $=11\times 100000=1100000$ cm.
  2. Step 2: Distance per revolution $=\frac{1100000}{5000}=220$ cm $=$ circumference.
  3. Step 3: $2\pi r=220\Rightarrow 2\times\frac{22}{7}\times r=220\Rightarrow\frac{44}{7}r=220$.
  4. Step 4: $r=220\times\frac{7}{44}=5\times 7=35$ cm.

Answer: Radius $=35$ cm.

9
Worked Example
If the perimeter and the area of a circle are numerically equal, find the radius of the circle.
Solution
  1. Step 1: Set the two numerical values equal: $2\pi r=\pi r^2$.
  2. Step 2: Divide both sides by $\pi r$ (since $r\neq 0$): $2=r$.
  3. Step 3: So the radius is $2$ units.

Answer: Radius $=2$ units.

10
Worked Example
A copper wire is bent into a circle of radius 28 cm. It is then straightened and re-bent into a square. Find the side and area of the square. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Length of wire $=$ circumference $=2\pi r=2\times\frac{22}{7}\times 28=176$ cm.
  2. Step 2: This becomes the perimeter of the square, so $4\times\text{side}=176$.
  3. Step 3: Side $=\frac{176}{4}=44$ cm.
  4. Step 4: Area of square $=44\times 44=1936\ \text{cm}^2$.

Answer: Side $=44$ cm and area $=1936\ \text{cm}^2$.

11
Worked Example
Find the area of a ring whose outer and inner radii are 23 cm and 12 cm. (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Ring area $=\pi(R^2-r^2)=\pi(R+r)(R-r)$ with $R=23$, $r=12$.
  2. Step 2: $=\frac{22}{7}\times(23+12)\times(23-12)=\frac{22}{7}\times 35\times 11$.
  3. Step 3: $=22\times 5\times 11=1210\ \text{cm}^2$.

Answer: $1210\ \text{cm}^2$.

12
Worked Example
The diameter of a wheel of a bus is 140 cm. How many revolutions per minute must the wheel make so the bus moves at 66 km/h? (Take $\pi=\frac{22}{7}$).
Solution
  1. Step 1: Radius $r=70$ cm. Circumference $=2\pi r=2\times\frac{22}{7}\times 70=440$ cm $=4.4$ m.
  2. Step 2: Speed $=66$ km/h $=\frac{66\times 1000}{60}=1100$ m/min.
  3. Step 3: Revolutions per minute $=\frac{1100}{4.4}=250$.

Answer: $250$ revolutions per minute.

Key Points

  • The perimeter of any circle is also called its circumference, calculated as $2\pi r$.
  • The area measures the inside flat space using the formula $\pi r^2$.
  • The constant Pi ($\pi$) is an irrational ratio value roughly equal to $22/7$ or $3.14$.
  • When a circular object rolls forward on the ground, the linear distance it travels in exactly one full spin equals its circumference.
  • If you double the radius of a circle, its perimeter doubles, but its area increases by four times ($2^2$).
  • To recast circles, equate the conserved quantity: areas add ($R^2=r_1^2+r_2^2$) or circumferences add ($R=r_1+r_2$).
  • A circular path of width $w$ around a circle of radius $r$ has area $\pi(R^2-r^2)$ where $R=r+w$.
Tap an option to check your answer0 / 4
Q1.The circumference of a circle is:
Explanation: $2\pi r$.
Q2.The area of a circle is:
Explanation: $\pi r^2$.
Q3.With $r=7$ and $\pi=\tfrac{22}{7}$, the area is:
Explanation: $\tfrac{22}{7}\cdot49=154$.
Q4.With $r=7$, the circumference is:
Explanation: $2\cdot\tfrac{22}{7}\cdot7=44$.