10. Modeling With Circles
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Chapter 5
10.
Modeling With Circles
This lesson explores the application of mathematical concepts like area calculations and trigonometric ratios to circular objects in real-life situations. For example, it helps you decide which pizza size offers more food by calculating the area. It also delves into how the Knights of the Round Table could determine the radius of their chairs using trigonometric ratios. These mathematical tools are not just theoretical but have practical applications in various aspects of life, such as food, furniture, and even timekeeping.
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Lesson Settings & Tools
| | 7 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Modeling With Circles
Slide of 7
Based on the features and theorems of the circles, this lesson will relate the circles with the real-life examples.
Catch-Up and Review
Here are some recommended readings before getting started with this lesson.
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Investigating Circular Shaped Objects
Tennis balls are sold in cylindrical containers that contain three balls each. The tennis balls and container can be modeled as follows.
The container, lying on its side, passes through an X-ray scanner in Tallahassee International Airport, where Jordan works as a security guard. Depending on the opacity of the container and the balls, what should Jordan expect to see when she reads the screen of the X-ray scanner? Explore the applet by adjusting the opacity of the container and the tennis balls.
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Modeling the Area of Circular Objects
Izabella and her friends are ordering pizza from their favorite spot. They can afford to buy either two 9-inch pizzas or one 13-inch pizza. The thickness of the pizza is the same for each.
They cannot quite decide which option means more food. By modeling each pizza as a circle, help Izabella and her friends determine the best option.
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Hint
Since both pizzas have the same thickness, the area of the pizzas will help to make a decision.
Solution
It is given that the pizzas have the same thickness. Therefore, areas of the pizzas will determine the best option. Begin by modeling the pizzas as circles with diameters of 9 and 13 inches.
| Radius | Substitution | Area | |
|---|---|---|---|
| 9-inch pizza | 4.5 | A=π ( 4.5)^2 | ≈ 63.62 |
| 13-inch pizza | 6.5 | A=π ( 6.5)^2 | ≈ 132.73 |
Recall that Izabella and her friends can afford two 9-inch pizzas. Area of two 9-inch Pizzas: ≈ 127.24 Area of one 13-inch Pizza: ≈ 132.73 As a result, the best option is to buy a 13-inch pizza. One of the friends was so close to ordering the two 9-inch pizzas. Thanks for helping them make the right choice.
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Modeling the Radius of Circular Objects
The Knights of the Round Table are the highly-esteemed knights of the mythical fellowship of King Arthur. In the beginning, there were only twelve knights. They met at the Round Table that is a symbol of equality among all of its legendary members.
Assume that each knight has the same circular chair around the Round Table whose radius is 10 feet. Given that all chairs and the Round Table are externally tangent to each other, what is the radius of each chair? Round the answer to the nearest tenth.
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Hint
Begin by modeling the Round Table and the chairs using circles. Draw a triangle whose vertices are the center of the inner circle and two adjacent outer circles. Then use the trigonometric ratios.
Solution
The Round Table and the chairs can be modeled by using circles.
By labeling the centers of the inner circle and two adjacent outer circles, a triangle passing through these centers can be drawn.
Segments that connect the centers of the tangent circles pass through the points of tangency. Assume that the radius of each chair is r feet.
Notice that △ ABC is an isosceles triangle where AB=AC. From here, to find the value of r, the vertex angle of △ ABC needs to be found. Since there are 12 chairs, the measure of ∠ BAC can be found by dividing 360^(∘) by 12. 360^(∘)/12=30^(∘) Now that the measure of the vertex angle was found, the altitude of the base can be drawn to use the trigonometric ratios. Remember that the altitude of the base of an isosceles triangle bisects the vertex angle and the base.
sinθ = BD/AB
SubstituteValues
Substitute values
sin 15^(∘) = r/10+ r
▼
Solve for r
MultEqn
LHS * (10+r)=RHS* (10+r)
10sin15^(∘)+rsin15^(∘)=r
SubEqn
LHS-rsin15^(∘)=RHS-rsin15^(∘)
10sin15^(∘)=r-rsin15^(∘)
FactorOut
Factor out r
10sin15^(∘)=r(1-sin15^(∘))
DivEqn
.LHS /(1-sin15^(∘)).=.RHS /(1-sin15^(∘)).
10sin15^(∘)/1-sin15^(∘)=r
RearrangeEqn
Rearrange equation
r=10sin15^(∘)/1-sin15^(∘)
UseCalc
Use a calculator
r=3.491981...
RoundDec
Round to 1 decimal place(s)
r≈ 3.5
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Modeling Bodies Using Cylinders
Jordan and Ali are comparing their weights. Jordan is 5 feet tall, and Ali is 6 feet tall. Their shoulder widths are 1 foot and 1.4 feet, respectively.
Consider their bodies to be roughly cylindrical, and their shoulder widths could represent the diameter. What is the ratio of Jordan's weight to Ali's weight?
Answer
4:9 or 4/9
Hint
Volume of a cylinder is the product of the base area and height of the cylinder.
Solution
When a body is modeled using a cylinder, one's height and shoulder width will represent the cylinder's height and diameter, respectively.
By comparing the volumes of these cylinders, a relation between the weights of Jordan and Ali can be set up. To do this, the radii of the cylinders should be found. Recall that the radius of a cylinder is half of its diameter.
Shorter Cylinder & Taller Cylinder 1/2=0.5 ft & 1.4/2=0.7 ft
Now that the radii have been found, the formula for the volume of a cylinder can be used to find their volumes.
| h | r | π r^2 h | V | |
|---|---|---|---|---|
| Shorter Cylinder | 5 | 0.5 | π ( 0.5)^2( 5) | ≈ 4 |
| Taller Cylinder | 6 | 0.7 | π ( 0.7)^2( 6) | ≈ 9 |
The volumes of the cylinders imply that if Jordan weighs 80 pounds, Ali weighs 180 pounds. Therefore, the ratio of Jordan's weight to Ali's weight is 4:9 or 49.
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Modeling Arc Measures
Arc measure is also an important concept to interpret the situations modeled by circles. For example, consider a time zone wheel.
A time zone wheel is a tool used to find the time in different locations across the world. For example, to find the time in Tokyo when it is 1P.M. in Amsterdam, rotate the small wheel until the labels of 1P.M. and Amsterdam align.
As can be seen, it is 9P.M. in Tokyo.
a What is the arc measure between each time zone on the wheel?
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b What is the measure of the minor arc from the Istanbul zone to the Los Angeles zone?
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Hint
a How many time zones are there on the wheel?
b How many time zones are there between the Istanbul zone and Los Angeles zone?
Solution
a It can be seen that the wheel is divided into 24 time zones. Since the measure of the whole wheel is 360 ^(∘), the arc measure between each time zone on the wheel can be found by dividing 360^(∘) by 24.
360^(∘)/24=15 ^(∘)
b Notice that the minor arc between the Istanbul zone and Los Angeles zone contains 11 time zones. Since the arc measure between each time zone is 15 ^(∘), the measure of the minor arc can be found by multiplying 15^(∘) by 11.
15^(∘) * 11 = 165 ^(∘)
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Modeling Real Life Situations Using Circles
In real life, there are plenty of situations where circles can be appreciated. In this lesson, a few common situations have been covered. How about some other fun ones? Take, for example, one of the most complex applications of circles — the global positioning system.
Modeling With Circles
Exercise 3.1
MacKenzie is making a water bowl made of clay for her little dog Scarlett.
Before shaping the bowl, she has sketched the design.
The part above the cylindrical base can be treated as a truncated cone. A truncated cone is a cone that has had a portion of the top cut off. For this particular truncated cone, MacKenzie removes a portion that is 8 centimeters tall. The bowl's carved-out part can be modeled by an upside-down truncated cone where a portion with a height of 56 centimeters has been removed.
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Let's first consider the shape of the clay before MacKenzie carves out the hollow. In this case, the bowl forms a solid block of clay consisting of a cylindrical plate with a truncated cone sitting on top.
Volume of Cylindrical Plate
From the diagram, we know that the cylinder has a radius of 16 centimeters and a height of 1 centimeter. With this information, we can determine the volume of this cylindrical plate V_c.
Truncated Cone
As described in the exercise, the design cuts away an 8 centimeter tall cone to create the 7 centimeter tall truncated cone. That means the full cone must have had a height of 15 centimeters. Additionally, since the cylindrical plate sticks out 1 centimeter from the cone, the full cone's radius is 15 centimeters. Similarly, the diagram shows that the cone that was cut away has a radius of 10 centimeters.
To determine the volume of this truncated cone V_(TC), we can subtract the volume of the cone that was cut away from the full cone. V_(TC)=1/3π (r_1)^2 h_1 - 1/3π (r_2)^2 h_2 Let's substitute r_1= 15 and h_1= 15 for the full cone and r_2= 10 and h_2= 8 for the cut-away portion.
Total Volume of Solid Block
By adding the volumes of the cylindrical plate and the truncated cone, we can determine the volume of the block of clay before carving out the hollow. V_C+V_(TC)=256π+1/3(2575π)= 1/3(3343π)
Volume of Upside-Down Truncated Cone
Like the first truncated cone, we will calculate the volume of the full cone and then subtract the top's volume. Examining the diagram and exercise description, we notice that the original upside-down cone has a radius of r_1= 9 and height of h_1= 63 as a full cone. The top portion that was cut away has a radius of r_2= 8 and height of h_2= 56.
Let's substitute these values into the formula for calculating the volume of the upside-down truncated cone.
Volume of Clay Needed
Finally, we can subtract the volume of the upside-down truncated cone, representing the hollow, from the solid block of clay. We will round our answer to the nearest integer. ( V_C+V_(TC))- V_(UTC) ⇓ 1/3(3343π)- 1/3(1519π)≈ 1910 cm^3
It will take 1910 cubic centimeters of clay to make Scarlett this water bowl.
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We can create two different cylinders from a rectangular sheet of paper depending on how the paper is rolled.
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Depending on how we roll the paper, the cylinder will either be short and wide or tall and narrow.
To determine each cylinder's volume, we will find their respective radius. We can do so by solving for the diameter in the formula for calculating a circle's circumference. C=π d Since we know the circumference for both cylinders, we can then solve for their diameter d.
| Cylinder | C=π d | Solve for d |
|---|---|---|
| Short | 20=π d | d=20/π |
| Tall | 10=π d | d=10/π |
When we know the diameter of each base, we get the radius by dividing each diameter by 2. Short Cylinder:& r_S=.20 /π./2=10/π [1em] Tall Cylinder:& r_T=.10 /π./2=5/π Let's now go ahead and calculate the volumes.
Short Cylinder
Tall Cylinder
Let's summarize the volumes. Short Cylinder:& V_S=1000/π cm^3 [1em] Tall Cylinder:& V_T=500/π cm^3 Since the short cylinder has a greater numerator, it is the cylinder with the greater volume.
If we divide the cylinder with the greater volume by the cylinder with the lesser volume we can determine the ratio.
The ratio is 2.
This time we want to determine the volume of an arbitrary paper with length a and width b. Remember, in this example a is greater than b.
In this case of rolling the paper, we get a short cylinder with a circumference of a and height b. We also get a tall cylinder with a circumference of b and height a.
Let's figure out the radius by using the formula for calculating a cylinder's circumference. Notice that the radius is the same thing as half the diameter, d2.
| Cylinder | C=π d | Solve for d/2 |
|---|---|---|
| Short | a=π d | d/2=a/2π |
| Tall | b=π d | d/2=b/2π |
Having found expressions for half the diameter, we can proceed with writing expressions for each cylinder's volume.
Short Cylinder
Tall Cylinder
Let's summarize the volumes. Short Cylinder:& V_S= a^2 b/4π cm^3 [1em] Tall Cylinder:& V_T= b^2 a/4π cm^3 Let's consider the numerators because the denominators are the same. Refer to the diagram and notice that a>b. Taking into account these two volume expressions, we can determine the short cylinder has the greater volume. Now we can divide the volume of the short cylinder by the volume of the tall cylinder to obtain the ratio.
The ratio is ab.
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Modeling With Circles
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