Circumference and Area of a Circle

The Circle — Circumference and Area

A circle of radius $r$ carries two basic measurements you have used since earlier classes.

• Circumference (the distance once around): $C = 2\pi r = \pi d$, where $d = 2r$ is the diameter.
• Area (the region enclosed): $A = \pi r^2$.

Here $\pi$ (pi) is the constant ratio of any circle's circumference to its diameter. For calculations we use $\pi = \dfrac{22}{7}$ (or $3.14$ when a question asks for it).

Key Point: $C = 2\pi r$ uses the radius once; $A = \pi r^2$ uses it squared. Mixing these up is the most common slip in the whole chapter.

From Diameter, and Back Again

Often a problem gives the diameter or the circumference, not the radius. Convert first:

• Given $d$: use $r = \dfrac{d}{2}$.
• Given $C$: use $r = \dfrac{C}{2\pi}$.

Worked idea

If the circumference is $44$ cm, then $2\pi r = 44 \Rightarrow r = \dfrac{44}{2\pi} = \dfrac{44 \times 7}{2 \times 22} = 7$ cm, so the area $= \pi r^2 = \dfrac{22}{7}\times 49 = 154$ cm$^2$.

[Board Important] When two circles are compared, remember: the ratio of circumferences equals the ratio of radii, but the ratio of areas equals the ratio of the squares of the radii.

Area of a Circular Ring (Annulus)

A ring (or annulus) is the region between two concentric circles — think of a washer, a circular track, or a pipe's cross-section. If the outer radius is $R$ and the inner radius is $r$, the ring's area is the big circle minus the small circle: $$\text{Area of ring} = \pi R^2 - \pi r^2 = \pi\left(R^2 - r^2\right).$$

Key Point: $R^2 - r^2 = (R-r)(R+r)$ — factorising often makes the arithmetic much easier.

A Rolling Wheel

When a wheel rolls without slipping, in one complete revolution it covers a distance equal to its circumference. So in $n$ revolutions it covers $$\text{distance} = n \times 2\pi r.$$

Turned around: number of revolutions $= \dfrac{\text{distance}}{2\pi r}$.

[Board Important] "How many revolutions does a wheel of radius $r$ make to cover $D$ metres?" is just $\dfrac{D}{2\pi r}$ — mind the units (convert everything to the same unit first).

Solved Examples

Example 1: Area from radius

Find the circumference and area of a circle of radius 7 cm. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. $C = 2\pi r = 2\times\dfrac{22}{7}\times 7 = 44$ cm.
2. $A = \pi r^2 = \dfrac{22}{7}\times 7^2 = 22\times 7 = 154$ cm$^2$.

Final Answer: $C = 44$ cm, $A = 154$ cm$^2$.

Takeaway: With $r=7$ (or a multiple of 7), $\pi=\tfrac{22}{7}$ cancels neatly.

Example 2: Radius from circumference

The circumference of a circle is 88 cm. Find its area. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. $2\pi r = 88 \Rightarrow r = \dfrac{88\times 7}{2\times 22} = 14$ cm.
2. $A = \dfrac{22}{7}\times 14^2 = \dfrac{22}{7}\times 196 = 616$ cm$^2$.

Final Answer: 616 cm$^2$.

Takeaway: Find $r$ first, then square it for the area.

Example 3: Area of a ring

A circular path is 7 m wide and its inner radius is 21 m. Find the area of the path. $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. Inner $r = 21$ m, outer $R = 21+7 = 28$ m.
2. Area $= \pi(R^2 - r^2) = \dfrac{22}{7}(28^2 - 21^2) = \dfrac{22}{7}(784-441) = \dfrac{22}{7}\times 343 = 1078$ m$^2$.

Final Answer: 1078 m$^2$.

Takeaway: Path/ring/track area $= \pi(R^2-r^2)$; outer radius $=$ inner $+$ width.

Example 4: Rolling wheel

A wheel of radius 35 cm makes 20 revolutions. What distance does it cover? $\left(\pi=\dfrac{22}{7}\right)$

Solution:

1. One revolution $= 2\pi r = 2\times\dfrac{22}{7}\times 35 = 220$ cm.
2. Distance $= 20 \times 220 = 4400$ cm $= 44$ m.

Final Answer: 44 m.

Takeaway: Distance $= (\text{revolutions}) \times$ circumference.