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JRLeon Geometry Chapter 6.5 – 6.6 HGSH
The Circumference / Diameter Ratio
JRLeon Geometry Chapter 6.5 HGSH
The distance around a polygon is called the perimeter.
The distance around a circle is called the circumference.
Recall that: Now:
Definitions
The Circumference / Diameter Ratio
Diameter
Radius
Center
JRLeon Geometry Chapter 6.5 – 6.6 HGSH
The Circumference / Diameter RatioWhich is greater, the length of three tennis-balls
or the circumference of one tennis-ball?
Wrap a string around the one ball to measure its circumference, then comparethis measurement with the length of the three tennis-balls.
SURPRISED?
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.
Let’s go investigate this relationship!
JRLeon Geometry Chapter 6.5 – 6.6 HGSH
The Circumference / Diameter RatioObject Circumference (C) Diameter (d) Ratio
1.
2.
3.
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 ).
By now you should be convinced that the ratio is very close to 3 for every circle.
We define as the ratio and the ratio is approximated as 3.14.
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 = 2r ) .
Compare your average with the averages of other groups. Are the ratios close?
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
• 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 of the ancient world. Archimedes showed the ratio to be between 3 and 3
• Mathematicians in ancient Egypt used ( as their approximation of circumference to diameter.
• Early Chinese and Hindu mathematicians used • By 408 A.D, Chinese mathematicians were using .• Today, computers have calculated approximations of to billions of decimal places!
A Little History
Circumference ConjectureIf C is the circumference and d is the diameter of a circle, then there is anumber such that C = __. If d = 2r where r is the radius, then C = ___.d 2r
C-65
JRLeon Geometry Chapter 6.5 – 6.6 HGSH
The Circumference / Diameter Ratio
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.
Example 1:
C = d The Circumference Formula Solution:
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
If a circle has circumference 12 meters, what is the radius?
Example 2:
C = 2r The Circumference Formula Solution:
12 = 2r 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 RatioJules 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.
Example 3 – Applications based on :
C = d The Circumference Formula Solution:C = (8000) Substitute the value of d, where d = 8000 milesC = (8000) C 3.14(8000) Substitute the value of 3.14C 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 = distance25,133 = r(80) rate x time = distance where d=25,133 miles and t=80 days
x = 13 mi./hr.
JRLeon Geometry Chapter 6.5 HGSH
The Circumference / Diameter RatioClasswork / Homework:
LESSON 6.5 : Pages 337-339, problems 1 through 15LESSON 6.6 : Page 342, problems 1 through 4.