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.
Explore
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.
Example
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.
9-inch Pizza29=4.5 in13-inch Pizza213=6.5 in
Now that the radii of the pizzas were found, the formula for the area of a circle can be used to find their areas. Begin by finding the area of 9-inch pizza.
The area of the pizza is about 63.62 square inches. Following the same procedure, the area of 13-inch pizza can also be found. | 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 |
Area of two 9-inch Pizzas: ≈127.24Area 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. Example
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.
12360∘=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=ABBD
SubstituteValues
Substitute values
sin15∘=10+rr
▼
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∘)
1−sin15∘10sin15∘=r
RearrangeEqn
Rearrange equation
r=1−sin15∘10sin15∘
UseCalc
Use a calculator
r=3.491981…
RoundDec
Round to 1 decimal place(s)
r≈3.5
Example
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 94
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 Cylinder21=0.5 ftTaller Cylinder21.4=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 | πr2h | 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 94.
Example
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 1 P.M. in Amsterdam, rotate the small wheel until the labels of 1 P.M. and Amsterdam align.
As can be seen, it is 9 P.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.
24360∘=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∘
Closure
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
Exercises
A new bakery is planning to make and sell delicious chocolate chip cookies. They use a 16-centimeter wide square sheet of cookie dough. The bakery can use two types of cookie cutters. One cutter cuts 16 cookies out of a sheet of cookie dough, and the other cuts 4 cookies out of an equally-sized sheet.
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We will determine how many small and large cookies the bakery can make with their cookie cutters to solve this exercise.
Small Cookies
Because each cookie has the shape of a circle, we can calculate its area if we know its radius. Since this cookie cutter can cut 4 cookies side by side along the width of the sheet, the diameter of each small cookie is 164= 4 centimeters.
At 4 centimeters in diameter, each small cookie then has a radius of 2 centimeters. We now have enough information to calculate the area of a small cookie.
One small cookie has an area of 4π square centimeters. Since the quantity of cookies this cutter can cut is 16, the combined area of the cookies is the product of 16 and the area of one small cookie. 16(4π) =64π cm^2 To determine how much dough is left over, we will subtract the area of the 16 cookies from the sheet's area. A_(sheet)- A_(cookies)= 16^2- 64π Finally, by dividing this difference by the area of one cookie, we can determine how many more small cookies we can make from the leftover dough.
We can make an additional 4 small cookies. That would make for a total of 20 small cookies from one sheet.
Large Cookies
As for the large cookies, we can fit 2 side by side on a sheet. That means each cookie has a diameter of 8 centimeters.
Therefore, each large cookie has a radius of 4 centimeters. Now we can calculate the area of one large cookie using the same method as before.
We will follow the same approach we did with the small cookies. We can find the area of the leftover dough the large cookie cutter initially cut by subtracting the area of a large cookie — multiplied by 4 — from the area of the sheet. A_(sheet)- A_(cookies)= 16^2- 4(16π) Dividing this difference by the area of one large cookie, we can determine how many more large cookies the bakery can make from the leftover dough.
Rounding to the nearest whole number, the bakery can make 1 more large cookie. That would make for a total of 5 large cookies that can be cut from one sheet.
Pricing of Large Cookies
From one sheet of dough, the bakery can make either 20 small cookies or 5 large cookies. Since a small cookie is sold for $1, the revenues from each sheet of cookie dough is $20. The bakery wants to make that same amount of money when selling large cookies. If we divide the revenue from the small cookies by the number of large cookies the bakery can make out of one sheet, we get the price the bakery should charge. $20/5=$4 The price for a large cookie should be $4.
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Benno the Clown got a brand new custom-made tent where he wants to relax between his performances. Unfortunately, the company that made Benno's tent got the measurements wrong. They made the tent so low that Benno has to crunch uncomfortably when he stands and walks in the tent.
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The tent can be modeled by a cone attached to the top of a cylinder. From the exercise, we have been given the height and radius of the cylinder. Since the cone is attached to the cylinder they will have the same radius.
We know that the volume of the tent is 5 cubic meters. Let's recall the formulas for calculating the volume of a cylinder and the volume of a cone. Cylinder:& V=π r^2 h Cone:& V=1/3π r^2 h Let's call the height of the cylinder h_1 and the height of the cone we call h_2. If we add the volume of the cylinder to the volume of the cone, the sum should equal 5 cubic meters.
By adding the height of cone to height of the cylinder we get an expression for the height of the tent h_t. h_t= h_1+ h_2 ⇓ h_t= 1.5 + 15-4.5π/π Since Benno's height h_B is given in centimeters, we need to convert it to meters. h_B=220 cm= 2.2 m Let's calculate the difference between Benno's height and the height of the tent.
This is how much higher Benno wants his tent to be. However, this is given in meters, we still need to convert it to centimeters. 0.42535... m= 42.535... cm When we round this to a whole number, the tent needs be 43 centimeters higher to fit Benno. Perhaps he can enjoy his break time a bit more now.
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The pharmaceutical Xanatrin is delivered in small capsules, as shown in the diagram.
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Examining the diagram, we notice that a capsule consists of two hemispheres with a cylindrical section between the two. When the two hemispheres are joined together, they form a sphere.
Therefore, to determine the total volume of the capsule, we have to add the volume of a cylinder and the volume of a sphere. Sphere:& V_S=4/3π r^3 [0.7em] Cylinder:& V_C=π r^2 h First, let's calculate the volume of the sphere.
Next, we will calculate the volume of the cylinder.
Now we can add the cylinder's volume to the sphere's volume to get the capsule's volume. 32π/3 + 40π = 152π/3 mm^3 Since each cubic millimeter of the capsule contains 3 milligrams of the active ingredient, we can find the total amount of Xanatrin by multiplying the volume by 3. 152π/3(3)≈ 478 mg
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You've been asked to organize the traffic at your friends dance recital next week. How many square centimeters of pliable plastic would you need to construct 100 hollow traffic cones like the one in the diagram?
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The traffic cone consists of a square shaped base with a hollow cone attached to it. We need to figure out the area of both components.
Area of Cone
Since the cone is hollow, it lacks a bottom. That means we are only considering the cone's surface area. That can be calculated using the following formula. S=π rl In this formula, r is the cone's radius and l is the length of its side. Since the diameter is 16 centimeters, its radius is 8 centimeters. We know that side length is 36 centimeters.
Area of Squared Base
The base is a square with a side length of 20 centimeters. Since the cone sits on the square, we have to subtract the area of a circular hole with a radius of 8 centimeters in order to get the surface area of the base.
Total Surface Area
Finally, we can add the area of the base and the area of the cone to get the total surface area TS. TS= 288π+ 400-64π ⇓ TS=( 400+224π) cm^2
Pliable Plastic Needed for 100 Cones
Since 100 traffic cones are being produced, we get the total amount of pliable plastic we need by multiplying the surface area by 100. 100( TS)=100( 400+224π) cm^2 ⇓ 100( TS)≈ 110 372 cm^2
You will need 110 372 square centimeters of pliable plastic. The traffic should flow well thanks to your help!
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Modeling With Circles
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