Circumference

What is pi?

You have probably heard of a special number in mathematics called Pi, it has the symbol $\pi$π. But what exactly is it? 

Well, Pi is a special number that represents the relationship between any circle's circumference and its diameter. 

Let's explore how many diameters fit around the circumference of a circle by using the applet below.  

• Change the diameter of the circle using the blue dot.  
• Then roll the circle out along the line.  
• Once the circle is unravelled you can lay the diameter segments end to end and see how many fit.  

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You would have found that the length of the circumference was always $3$3 and a bit diameters. In fact, this value is ALWAYS the same, and we call this constant Pi. 

The number Pi is irrational (and so the decimals go on and on forever, without ever repeating), that's why we use the symbol $\pi$π.  

$\pi=3.141592653589$π=3.141592653589...

If a number is irrational, then we cannot express it as the ratio of two whole numbers. Because Pi is irrational, any calculation we do on a calculator with Pi in its decimal form will be an estimate. A pretty close one, but still not what we would call exact. When we want to write an expression in "exact form", we can write it in terms of some multiple of $\pi$π.

It doesn't matter what the diameter is - the relationship between the circumference and diameter is always the same.  

$\pi=\frac{C}{d}$π=Cd​

Circumference

The circumference of a circle can be calculated with a special formula. 

To find the formula for the circumference of a circle we will rearrange the ratio we used to find $\pi$π and make $C$C the subject.

 

Practice Questions

QUESTION 1

QUESTION 2

QUESTION 3

Finding radius or diameter

Finding the radius of a circle using its circumference

We have the equation $C=2\pi r$C=2πr which gives the circumference of a circle in terms of its radius. Let's now rearrange this equation to make the radius the subject.

Dividing both sides of the equation by $2\pi$2π, we are left with

$r=\frac{C}{2\pi}$r=C2π​.

So, substituting the circumference $C$C into the equation $r=\frac{C}{2\pi}$r=C2π​ will give the exact value for the radius $r$r.

 

Finding the diameter of a circle using its circumference

Another related part of the circle is the diameter $D$D, which is twice as long as the radius $r$r. That is,

$D=2r$D=2r.

This means that we can also find the diameter of a circle when we know its circumference. Since we know that the radius is given by the equation

$r=\frac{C}{2\pi}$r=C2π​,

and that the diameter is twice as long as the radius, this means that

$D=2\times\frac{C}{2\pi}$D=2×C2π​.

Cancelling the common factor of $2$2 between the numerator and denominator, we have that

$D=\frac{C}{\pi}$D=Cπ​.

So, substituting the circumference $C$C into the equation $D=\frac{C}{\pi}$D=Cπ​ will give the exact value for the diameter $D$D.

 

 

It is sometimes more useful to know the radius or diameter of a circle as a decimal, rather than the exact value in terms of $\pi$π. If this is the case, we can use a calculator to evaluate the expression and round the result to the desired number of decimal places.

 

Practice questions

Question 4

Question 5

Finding the fractional part of the circumference

Fractional parts of circles are often referred to as sectors. These are effectively pieces of a circle that have been sliced out from the centre.

examples of sectors

 

If we wanted the circumference for just $\frac{1}{4}$14​ of a circle, we would find the circumference and divide by $4$4.  

If we wanted the circumference for just $\frac{1}{3}$13​ of a circle, we would find the circumference and find 1 third of it.  

So the process is going to be the same as above, except we will also have to find the fractional part of the circumference value. 

 

Practice Questions 

Question 6

Question 7

Problem solving with circular shapes

If you take a look around, you should be able to spot some objects that have circular features. A clock face, a round plate, or the lid of a jar are some examples of things that have the outline of a circle.

For lots of other common objects there are circles present in different parts of their shape. Some examples include the rim of a cup, or the crust on a pizza.

Once we have found a feature of an object that looks like a circle, we can explore the relationship between the radius, the diameter, and the circumference of that circle. In his way we can understand more about the overall object.

 

 

Worked example

Lisa is cleaning the leaves out of the pool in her backyard. The pool is a circular shape and has a radius of $5$5 m. What distance does Lisa cover if she walks all the way around the pool? Give your answer to one decimal place.

Think: The distance around the outside of a circle is its circumference. We can use the radius of the pool to find its circumference.

Do: Using the equation $C=2\pi r$C=2πr, we substitute $r=5$r=5 m.

Now we have $C=2\pi\times5=31.41592$C=2π×5=31.41592$...$... m, which rounds to $31.4$31.4 m. So Lisa will walk $31.4$31.4 m around the pool.

Reflect: Suppose Lisa's neighbour has a circular pool with twice the circumference. What would be its radius?

 

Circular motion

We can also use these same ideas about circles to understand the circular motion of objects. Think about the way the end of the second hand moves around a clock, or the way a wheel rolls down a hill.

Worked example

Harry pushes a hula hoop down a hill, and it rolls all the way to the bottom. The hula hoop has a diameter of $85$85 cm. If the hula hoop rolled a total of $25$25 m, how many revolutions did it make on its way down the hill? Give your answer to two decimal places.

Think: The hula hoop will make one full revolution when it rolls a distance equal to its circumference. We can use the diameter to find the circumference, then divide the total distance the hoop rolled by this length to find the number of revolutions.

Do: Using the equation $C=\pi D$C=πD, we find that the circumference is $85\pi$85π cm. So the hoop will roll $85\pi$85π cm in one revolution.

Since the total distance is given in metres, we will convert the circumference to the same units. There are $100$100 cm in $1$1 m, so the circumference $85\pi$85π cm is equal to $\frac{85}{100}\pi=0.85\pi$85100​π=0.85π m.

Now we can divide the total distance rolled by this circumference.

Number of revolutions	$=$=	$\frac{\text{total distance rolled }}{\text{circumference }}$total distance rolled circumference ​
	$=$=	$\frac{25}{0.85\pi}$250.85π​
	$=$=	$9.36$9.36 (to two d.p.)

The hula hoop has rolled about $9.36$9.36 times as it covers the distance of $25$25 m.

Reflect: How far would the hula hoop have rolled if it completed exactly $11$11 revolutions?

 

Practice questions

Question 8

Question 9

Question 10