The Circumference / Diameter
Ratio
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Recall that:
Now:
The distance around
a polygon is called
the perimeter.
The distance around
a circle is called the
circumference.
Diameter
Center
Definitions
JRLeon
Geometry Chapter 6.5
HGSH
The Circumference / Diameter
Ratio
Which is greater, the length of three tennis-balls
or the circumference of one tennis-ball?
SURPRISED?
Wrap a string around the one ball to measure its circumference, then compare
this measurement with the length of the three tennis-balls.
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Now that you actually compared the measurements, you discovered
that the circumference of the tennis-ball is greater than three diameters
of the ball.
In this lesson you are going to discover the relationship between the
diameter and the circumference of every circle.
Once you know this relationship, you can measure a circle’s diameter
and calculate its circumference.
JRLeon
Geometry Chapter 6.5 - 6.6
HGSH
The Circumference / Diameter
Ratio
Object
Circumference (C)
Diameter (d)
Ratio
𝑪
𝒅
1.
2.
3.
𝐶
Compare your average with the averages of other groups. Are the 𝑑 ratios close?
𝐶
By now you should be convinced that the ratio 𝑑 is very close to 3 for every circle.
𝐶
We define  as the ratio 𝑑 and the ratio is approximated as 3.14.
If you solve this formula for C, you get a formula for the circumference of a circle in terms
of the diameter, d ( C = d ).
The diameter is twice the radius (d =2r), so you can also get a formula for the
circumference in terms of the radius, r ( C = 2r ) .
If you’re asked for an exact answer instead of an
approximation, state your answer in terms of 
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
C-65
Circumference Conjecture
If C is the circumference and d is the diameter of a circle, then there is a
d If d = 2r where r is the radius, then C = ___.
2r
number such that C = __.
A Little History
• The first calculation of the circumference to diameter ratio, known as pi ( ), was
done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians
1
10
of the ancient world. Archimedes showed the ratio to be between 3 7 and 3 71
4
• Mathematicians in ancient Egypt used (3)4 as their approximation of circumference
to diameter.
• Early Chinese and Hindu mathematicians used 10.
355
• By 408 A.D, Chinese mathematicians were using 113.
• Today, computers have calculated approximations of  to billions of decimal places!
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Example 1:
If a circle has diameter 3.0 meters, what is the circumference?
Use a calculator and state your answer to the nearest 0.1 meter.
Solution:
C = d
The Circumference Formula
C = (3.0)
Substitute d with 3.0
In terms of , the answer is 3 . The circumference is approximately 9.4 meters
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Example 2:
If a circle has circumference 12 meters, what is the radius?
Solution:
C = 2r
The Circumference Formula
12 = 2r
Substitute the value of C
6= r
Substitute the value of C
So the radius is 6 meters.
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Example 3 – Applications based on  :
Jules Verne’s “Around the world in 80 days”
If the diameter of Earth is 8000 miles, find the average speed in miles per hour
Phileas Fogg needs to circumnavigate Earth about the equator in 80 days.
Solution:
C = d
The Circumference Formula
C = (8000)
Substitute the value of d, where d = 8000 miles
C =  (8000)
C  3.14(8000)
Substitute the value of   3.14
C  25,133 miles
So, Phileas must travel 25,133 miles in 80 days. To find the speed v in mi/h, you need to
divide distance by time and convert days into hours.
d = rt
rate x time = distance
25,133 = r(80)
rate x time = distance where d=25,133 miles and t=80 days
𝟐𝟓,𝟏𝟑𝟑 𝒎𝒊𝒍𝒆𝒔
𝟏 𝒅𝒂𝒚
x
= 13 mi./hr.
𝟖𝟎 𝒅𝒂𝒚𝒔
𝟐𝟒 𝒉𝒐𝒖𝒓𝒔
JRLeon
Geometry Chapter 6.5 – 6.6
HGSH
The Circumference / Diameter
Ratio
Classwork / Homework:
LESSON 6.5 : Pages 337-339, problems 1 through 15
LESSON 6.6 : Page 342, problems 1 through 4
.
JRLeon
Geometry Chapter 6.5
HGSH