NZ Level 7 (NZC) Level 2 (NCEA)
Circumference and Area of Circles in Exact Values
Lesson

We are told in geometry lessons that there is a fixed ratio between the circumference and the diameter of a circle. We write $\frac{C}{D}\approx\frac{22}{7}$CD227. In other words, $C\approx\frac{22}{7}D$C227D.

You may wonder how we can be certain that this ratio is the same for circles of every size. Think of some similar triangles, one small and one larger.

The ratio $\frac{A}{B}$AB in the small triangle is the same as the ratio $\frac{a}{b}$ab in the larger triangle. Now imagine piecing together enough copies of the small triangle, side by side, to make a polygon approximating a circle with radius $B$B. The same can be done with the larger triangle, to make a similar polygon with radius $b$b.

Suppose $k$k copies of each triangle were needed to make the polygons. Then, in the smaller polygon, the ratio of the perimeter to the length $B$B must be $\frac{kA}{B}$kAB. In the larger polygon, the same ratio would be $\frac{ka}{b}$kab. These ratios must be the same because $\frac{A}{B}=\frac{a}{b}$AB=ab.

Now imagine doing the same procedure but with much thinner triangles. The polygons would still have the same ratio $\frac{A}{B}=\frac{a}{b}$AB=ab no matter how thin the triangles were, but with thin enough triangles the polygons would be indistinguishable from two circles. This suggests that the ratio of circumference to radius and hence, circumference to diameter does not change when the size of the circle changes.

The ancient Greek mathematician Archimedes used a similar kind of argument involving a circle with inscribed and circumscribed polygons to deduce that the ratio $\frac{C}{D}$CD is between $3\frac{10}{71}$31071 and $3\frac{10}{70}$31070.

Thus, the exact value of the ratio $\frac{C}{D}$CD is very slightly less than $\frac{22}{7}$227. We now know that it is an irrational number, meaning it cannot be written down precisely as a decimal or any other kind of fraction. We give it the symbol $\pi$π.

We can approximate $\pi$π to any level of precision required for practical purposes. To five decimal places, it is $3.14159$3.14159, for example. But when we wish to be precise, we just use the symbol $\pi$π.

The area of a circle is also connected with the number $\pi$π

One can imagine a circle cut into a large number of thin sectors. The sectors can be rearranged to form a shape approaching a rectangle. One side of the rectangle is the radius $r$r of the sectors and the other edge is approximately half the circumference of the circle. Thus, the area of the rectangle approaches $r\times\frac{\pi D}{2}$r×πD2 which is just $\pi r^2$πr2

The derivation of the area of a circle is made rigorous using integral calculus.

So, we use the circumference formula

$C=2\pi r$C=2πr

and the area formula

$A=\pi r^2$A=πr2

#### Example 1

The earth travels around the sun in an approximately circular orbit with average radius $149600000$149600000 km. How far relative to the sun does the earth travel in a year?
What is the area of the imaginary disc formed by the earth's orbit around the sun?

In $1$1 year the orbit covers the circumference of a circle with radius $149600000$149600000 km. So, the distance is about $C=2\times\pi\times149600000\approx939964522$C=2×π×149600000939964522 km. That is, about $9.4\times10^8$9.4×108 km. As an exact value, we could say $C=299200000\pi$C=299200000π but this is not really 'exact' due to the measurement error in the radius of the orbit.

The orbit encloses an imaginary disc with area $A=\pi\times149600000\times149600000\approx70310000000000000$A=π×149600000×14960000070310000000000000 km$^2$2. That is, $7.03\times10^{16}$7.03×1016 km$^2$2.

#### Example 2

The circumference at the greatest circular cross-section of an ordinary chicken egg is measured to be $151$151 mm.

What is its diameter and what is the area of the cross-section at the widest point?

Using $C=\pi D$C=πD, we have $151=\pi D$151=πD. Therefore, the diameter is $D=\frac{151}{\pi}\approx48$D=151π48 mm.

From this, the radius must be $24$24 mm and hence, the cross-sectional area is $A=\pi\times24^2\approx1809$A=π×2421809 mm$^2$2. This is $18.09$18.09 cm$^2$2.

#### Worked Examples

##### Question 1

If the radius of a circle equals $15$15 m, find its circumference.

##### Question 2

The circumference $C$C of a circle is equal to $264$264 m and its radius is $r$r m long.

Find the value of $r$r.

##### Question 3

Find the exact area of the circle shown.

Give your answer as an exact value in terms of $\pi$π.

##### Question 4

Find the shaded area of the figure shown.

### Outcomes

#### M7-4

Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions

#### 91259

Apply trigonometric relationships in solving problems