• Lesson ideas

    There are three types of lesson ideas.  Open-ended projects are major activities that engage students in answering non-trivial questions about the real-world contexts.  These activities could be treated as projects that take several days or class meetings.  Free-response items are individual problems or sets of problems. They require students to make decisions, calculate values, and do other things to construct their own answers.  For all or at least most of the open-ended items, there are specific suggestions about how the items may play out in the classroom.  Variety items are stand-alone problems that could be used in many different ways, including review and practice.  These items may be used to revisit previously learned materials while students are working either on a project or on several open-ended items related to cell phones.


    Open-ended project

    P1. Tower Power

    A cell phone will work when you are close enough to a cell phone tower.  A cell phone tower is a tall structure, like the one in the photo here.


    A cell phone company needs to pay money to set up a tower.  The company also pays money to keep the tower working.  So, a cell phone company wants to have only as many towers as it really needs and to have the towers in places where the most people will use them. It probably does not surprise you to know that cell phone towers are often located near towns and cities, places with large numbers of people.

    Here is a map of Dauphin County.  On the map we have marked several places with large numbers of people. 

    A cell phone tower typically can be used by people who are at most one or two miles from the tower.  The “tower tool” shown here may be helpful to you in measuring one or two miles from the tower.  The shorter plastic “arm” represents one mile on the map.  The longer plastic arm represents two miles on the map.

    a.     If a cell phone tower covers a one-mile radius, where would you locate the towers?  Explain your thinking.

    Answer: Guess and test is possible, trying to include as many towns as possible with each tower. The blue dots in the following map represent one possible solution.

    b.     If a cell phone tower covers a two-mile radius, where would you locate the towers? Tell how you arrived at this answer

    Answer: One possible solution is to place the towers where there are red dots on this map. Note the mark in the “350” for Berrysburg.  This point is barely within 2 miles of all the towns in the north.

    c.     Compare the numbers of towers you have as answers in parts a and b.  Which number of towers is larger?  Why does this make sense?

    Answer:  The number of towers for part a is larger than the number of towers for part b.  This makes sense because each circle with radius 1-mile is smaller than each circle with the 2-mile radius.  The area of each smaller circle is π≈3 square miles. The area of each larger circle is 22π≈4(3)=12 square miles.  That means each larger circle covers roughly 12÷3 or approximately 4 times as much land as each smaller circle.

    Enactment:  Have students compute the relative areas and compare the percentages.  This work not only includes more experience with circle area and percent, but it also leads to the idea that the ratio of the areas (4:1 for the circles) is the square of the ratio of the linear distances (2:1 for the radii).


    P2.  Two people can use cell phones within feet of each other but each person hears only their own phone call.  Why is it that cell phones seem “smart enough” not to confuse one call with another call?  The answer involves how digital information is coded and transmitted.

            The Penn State group would really appreciate feedback on this project.  Please send your thoughts to Rose at rmz101@psu.edu.  This project involves some simple probability and some tree diagrams.  Is this something worth including in the module?


    Free-response items

    F1. Building Towers

    Here is a map of Dauphin County.  On the map, we have marked several places with large numbers of people.  To decide where to place towers, the company divides the county into a north part and a south part, as shown by the thick red line.  Would the company have more customers in the north part of the county or in the south part of the county?  Explain your reasoning.


    Way : Add the populations in the north to get 9,346.  Add the populations in the south to get 76,797.  76,797 is greater than 9,346.  There are more people in the south.

    Way : Put the populations from the north in order from least to greatest.  Put the populations from the south in order from least to greatest.  Matching up corresponding populations, we see all of the numbers in the south are larger than the corresponding values in the north.  The south has more people.

    Way ƒ:  Estimate the sum in the north as 13000 and note that this total is less than the population of Harrisburg.  There are more people in the south.


    F2. Exploring Our County

    Here is a map of Dauphin County.  The scale on the map is 1 in = 3/4 mi.  Name all of the places that are more than five miles from Harrisburg.  Explain how you arrived at your answer.

    Answer: The places are Middletown, Royalton, Hummelstown, Halifax, Millersburg, Elizabethville, Berrysburg, Pillow, Garte, Williamstown, and Lykens.  To find the solution multiply by 5 miles.  5 miles x 1 inch/(3/4) mile = 5(4/3) inch = 20/3 inches = 6 2/3 inches.  Then locate all places more than 6 2/3 inches from Harrisburg.




    F3. Exploring More of Our County

    Here is a map of Dauphin County.  The scale on the map is 1 in = 3/4 mi. How many places on the map are within 10 miles of Harrisburg

    Answer: To find the number of inches that represents 10 miles: 10 miles x 1 inch/(3/4) miles = 10(4/3) inches = 40/3 inches = 13 1/3 inches ≈ 13.33 inches.  Find all of the places within 13.33 inches of Harrisburg.  There are five places, including Steelton and Highspire.




    F4. Where Have They Gone

    In 2002 the population of Steelton Borough was 5,796 and the population of Highspire was 2,688.[1]  The population of Steelton Borough has been decreasing lately at the rate of -1.1% every two years.   The population of Highspire Borough has been decreasing at the rate of -1.2% every two years. If these trends continue, will there ever be a time when the populations of Steelton and Highspire are equal?

    Answer ideas:

    Way : Write linear functions, graph, and look for intersection but be unable to find one.

    Way : Write linear expressions, set them equal, attempt to solve, find there is no real solution.

    Way ƒ: Observe that Steelton is already bigger and loosing people more slowly than Highspire is loosing people. 

    Enactment: If students do not suggest Way ƒ, mention it.  Have students take this observation and explain how the graphs of Way  show the observation in Way ƒ.



    F5. Predicting Populations

    In 2002, the population of Highspire Borough was 2,688.  In 2000, the population of the Borough was 2,720. [2]  Assume the Borough’s population continues to follow a linear pattern and answer the following questions.

    a.     What was the population in 2001?


    Way : Use (2000,2720) and (2002,2688) as two points to find the slope of a line.  Then, substitute x=2001 and solve for y to determine the population in 2001.  (2720-2688)/(2000-2002)=–16.

     (2720 – y)/(2000-2001) = –16;  (2720-y)=16;  –y=-2704; y=2704.

    Way :  This is similar to Way  but it uses a different scale for the year.  Suppose time 0 is the year 2000 and time 2 is the year 2002.  Use the points (0,2720) and (2,2688) to determine the equation of the line.  Substitute x=1 to obtain 2720-y=-16; y=2704. This suggests the population in 2001 would be approximately 2704.

    Way ƒ: 2001 is the average of 2000 and 2002.  Find the average of 2720 and 2688.  This average is (2720+2688)/2 = 5400/2 = 2704, which is the expected population in 2001.

    b.     What would be the expected population of the Borough in 2005?


    Way : Use the same linear equation as in Way  of part a.  Substitute x=2005 to get an estimated population of  (2720-y)/(2000-2005)=–16; 2720-y=-16(-5); 2720-y=80; -y=-2640; y-2640.

    Way : Use the linear equation from Way of part a.  Substituting x=5 yields a population estimate of (2720-y)/5=-16; 2720-y=5(-16); 2720-y=-80; -y=-2640; y= 2640.

    Way ƒ:  From 2000 to 2002 the population decreased by 32.  From 2002 to 2004 the population should decrease again by 32.  The 2004 population would be –(32+32) + 2720 = 2656.  But, from 2004 to 2005, the population should decrease by 16, so 2654-16=2640 is the population.  The population of the Borough in 2005 would be –(64 +16) + 2720 = 2640.

    Enactment:  Have students think about why Way ƒ works for part a but not for part b.  Ask them also for what years that method would work if students in future were given the population information for 2004 and 2010.

    Have students explain how slope of linear functions relates to the decreases in population, a decrease of 32 from 2000 to 2002 and from 2002 to 2004 as well as a decrease of 16 from 2004 to 2005 and then a decrease of 80 from 2000 to 2005.  The answer could note a constant decrease of 16 people per year.




    F6. Reaching Out

    A tower is placed at a boundary of Dauphin County.  The placement of the tower would be the center of the circle on this map.  The radius of the circle represents the 2-mile range of the tower. What is the area of portion of the tower's range that is in Dauphin County?


    Way : Use 3 purple fraction-circle pieces to 'fill in' the area outside the circle.  This is 3/10.  Inside the county would be 1-3/10=7/10 of the region.  7/10*4π square miles is (14/5)π=2-4/5π≈8.8 square miles. 

    Way : Estimate obtuse angle is approximately 120° or 1/3 of 360°.  So reflex angle measures approximately 2/3 of 360°.  The area of the portion inside the circle would be approximately 2/3 of 4π square miles or (2/3)(22π)=(8/3)π≈8.4 square miles.

    Enactment:  Compare the answers in Way  and Way .  Ask students why the numerical answers – 2-4/5π≈8.8 square miles and (8/3)π≈8.4 square miles are close but the answer from Way  is greater than the answer to Way .  The discussion should include two ideas.  First, the estimate in Way  (3/10) is less than the estimate in Way (1/3).  Second, subtracting the larger estimate in Way leads to a smaller answer for Way than for Way .  This last notion may be counter-intuitive.




    F7. Service Area

    A cell phone tower provides service to customers within a particular region.  An example of the region that might be covered by one tower appears below. 

    To make its customers happy in a populated area like the Steelton-Highspire community shown in the map below, the company wants to cover every part of the community.


    a.     If every cell tower served a region like that in the picture, would it be possible to locate towers to cover all parts of the community?  Explain.

    Answer:  It would be possible to cover the community but there would be overlap.

    b.     Cell phone companies can think about the coverage of a tower as a simple circle, such as the one shown here.  If the tower regions are circles, determine how you would place the towers in order to cover the entire community.


    Way : Line up radii along parallel and perpendicular lines.  There is considerable overlap.

    Way : Line up radii along intersecting lines.  There is less overlap.


    Compare the overlap in both cases.  Talk about why the overlap varies like this.

    c.     In order to think about how to place towers, cell phone companies often think about the region each tower covers as a regular hexagon, as shown here. Is it possible to cover the community with regions of this shape? Explain.

    Answer: Yes.  The covering would look like this:

    d.     In each of parts a, b and c, you created a covering of a region using a geometric shape.  Imagine that you continued each pattern to try to cover the entire plane. Which, if any, of these extended patterns would be a tessellation?

    Answer:  Only the design in part c is a tessellation.  The other two options have overlaps, uncovered regions, or both.

    e.     What is the measure of an interior angle of a regular hexagon?  How does knowing this angle measure help to explain why the design in part c leads to a tessellation?  Explain your answers.

    Answer for measure of interior angle:

    Way : The measure of an interior angle of a regular hexagon is 120°.  We can find this through an estimate using Geometer’s Sketchpad or Cabri Geometry to create and measure several sample regular hexagons.

    Way : Draw the diagonals that connect ‘opposite’ vertices.  These meet at a point (the “center”) to create six triangles. Each angle of the triangle at the center is 360°/6=60°.  Since the distance from the center to a vertex is the same for every vertex, each of the triangles must be isosceles – in fact, equilateral, and so all of the angles of the triangles measure 60°.  The measure of any interior angle of the regular hexagon is the sum of the measures of two interior angles of the triangles.  So, the measure of an interior angle is 2(60°)=120°.

    Way ƒ: Using the formula 180°(n-2)/n for the measure of the interior angle of a regular n-gon, we have n=6 and the angle measure is 180°(6-2)/6=30°(4)=120°.

    Answer to how this relates to a tessellation:

    A whole number of hexagons have to meet at their vertices in order to have a covering.

    f.      Use your observation about interior angle measures in part c to predict whether a regular octagon will tessellate.  Explain your thinking.

    Answer:  The interior angle of a regular octagon is 180°(n-2)/n with n=8.   This measure is 180°(8-2)/8=180°(3/4)=45°(3)=135°.  135 does not divide 360.  So, I would not expect the regular octagon to tessellate.

    Enactment: It may help for students to have the formula in Way ƒ of part e to compute the interior angle measure. If not, they may measure a sample paper or electronic regular octagon to get an estimated measure.  Note that the inability of the regular octagon to tessellate is a good reason why the cell phone companies would not use a regular octagon as the model for the region covered by one tower in a grid of towers.





    F8 – Cell Tiles

    To illustrate the area covered by cell phone towers, engineers often use a grid like the following:

    All of the polygons are regular hexagons that are the same size and shape.  The length of each side of a hexagon is 2 miles. What is the area of one hexagon?


    Way : Draw a hexagon to scale using graph paper.  First sketch one side.  Measure angles of 120° as marked in the figure below.

    Draw a line through the endpoints of the new segments as shown.  Reflect the hexagon over that line as shown here.


    Count the grid boxes as shown in the last figure to the right. There are six boxes completely inside the hexagon (as shaded in yellow).  There are two parts of boxes at the top (both shaded in blue); each of these parts is almost one-half of a box.  The rest of the hexagon is made of four triangular regions.  Each of the triangles seems to cover close to half of two boxes, making the area of each triangle approximately half of 2 or 1 square mile.  So, the area of the hexagon is approximately  square miles.

    Way : Move one of the marked triangles (1) to the upper right. Move the other marked triangle (2) to the lower right as shown here:

    The hexagon has the same area as this rectangle that is 3 miles wide and 3.5 miles high.  The area of the rectangle is approximately 3x3.5=10.5 square miles.

    Way ƒ: The hexagon consists of six triangles, each with a vertex at the center of the hexagon and a side that is a side of the hexagon. 

    DABC is one such triangle.  The measure of ÐBAC is 360°÷6=60° and AB=AC, which is enough to conclude DABC is an equilateral triangle.  So, AB=AC=2 miles.  If M is the midpoint of segment BC, then MB=2÷2=1 mile.  AM is  miles. The area of DAMB is square miles, or approximately  square miles.  The area of the hexagon is 12 times the area of one of these triangles, or , approximately  square miles.

    Way : Tear the hexagon into 6 triangles (as shown to the left) and line them up (as shown to the right). 

    The result is a parallelogram approximately 6 miles wide and 1.7 miles high.  The area of the parallelogram is approximately 6x1.7=10.2 square miles.

    Explanation of differences and similarities:  All ways involve thinking of the hexagon as a collection of familiar figures. Some of the ways require measuring lengths or areas with standard units while others depend on the Pythagorean Theorem to compute an unknown length.  Some methods involve knowing the central and interior angles of a hexagon measure 60° and 120°, respectively.


    Enactment: Access to rulers, paper, grid paper, and scissors make it possible to try most of these solutions. Students are not expected to know the formula for the area of a hexagon.  From either Way ƒ  or Way , students may develop the formula, or at least see how the formula makes sense by looking at a general form of the particular calculations:

    Way ƒ:  


    Way : 



    F9 – Cell Area Shapes        

    A cell phone tower can be used only within a certain distance from the tower.  As a simple model of the region that a tower is useful, engineers use a hexagon like the one on the left. In actually, however, weather and other things get in the way.  The actuality region in which a cell phone tower may be useful often looks more like the shape on the right.  What is the area of this unusually shaped region?


                                                                          ç  4  miles  è


    Way : The area is less than 16 square miles.  The region is completely contained in a square with a side of length 4 miles.

    Way : The shape is similar to the polygon in the following figure with estimated measures as shown.

    The area of the region is the area of the square less the areas of the four shaded regions.  So, the area is square miles.

    Way ƒ: Place a 1/2-mile grid over the region as shown, with each square representing 1/4 square mile.  The region includes the 25 full squares (yellow), 14 half-squares (shaded red), the 4 quarter-squares (green), and a little bit more that may be at least 1 quarter square. The region covers approximately  squares. So, the area of the region is approximately  square miles.


    Explanation of differences and similarities: Notice that Way and Way ƒ are similar in that they are estimates.  In fact, Way  is very similar to Way ƒ but subtracting the total area of the shaded region from the full 16 square miles. The shaded region includes 15 full squares (yellow), 6 half-squares (red), and 4 triangles (orange) with areas each equivalent to a full square.  There are  squares shaded.  The unshaded squares, representing the region are 64-22=42 squares.  So, the estimated area of the region is square miles.

    Enactment: Encourage students to look at the different methods and discuss why there are such major differences among the numbers.  For another experience in approximating area, consider the items involving the map and populations of Dauphin County.

    [2] These data may be found at http://www.census.gov/popest/cities/tables/SUB-EST2003-04-42.pdf .


    F10. Cell Phone Expense    

    A Nextel plan costs $39.99 per month.  That cost includes 500 minutes.  Each additional minute costs $0.40.  Long-distance minutes cost $0.20 extra.  Tad received his monthly Nextel bill in the amount of $59.99.  How is that possible?  Explain your thinking.


    Way ¬: The $59.99 includes the per month charge of $39.99.  That leaves $59.99-$39.99=$20 that I need to account for.  The extra $20 might be $20/$0.20=100 long-distance minutes.

    Way ­: The extra cost of $(59.99-39.55)=$20 could be the result of $20÷$0.40 per minute =50 overtime minutes.


    Have students talk about how any number of overtime minutes and any number of long-distance minutes has to be less than the answers to Way ­ and Way ¬, respectively.


    F11. Tower Talk

    Jeanne and John were talking about the towns near Harrisburg and where cell phone towers might be placed.  They consulted the map shown here. As they talked, Jeanne said that there were exactly 4 towns that were seven miles from Harrisburg.  John, on the other hand, said that there were 5 towns that were seven miles from Harrisburg.  Who is correct and why?

    Answer: John is correct.  7 miles x 1 inch/(3/4) mile = 7(4/3) = 28/3 = 9 1/3 ≈ 9.3 inches. There are 5 towns within 9.3 inches of Harrisburg.


    F12.            Birthday Surprise

    Dominick celebrated his 13th birthday with a big party at his grandmother’s house in Elizabethville.  The party was great fun, but the best surprise came when it was time for the party to end.  Dominick’s grandmother had a hot-air balloon waiting to take him and his family back to Harrisburg. Although Dominick was very excited, he was also a little nervous. He said, “Grandma, how can we ride in a hot-air balloon for over 10 miles?”  His grandmother responded, “Don’t worry honey, it’s less than 10 miles. You’ll be fine.” Do you agree with Dominick or with his grandmother?  Explain your choice.

    Answer: Grandma is correct.  The distance on the map from Elizabethville to Harrisburg is approximately 13 inches.  13 inches x (3/4) mile/ 1 inch = 39/4 = 9 3/4 or 9.75 miles.

    Enactment: Students may reason that the distance is 10 miles by rounding.  This could lead to a discussion on the notion of rounding and its importance at arriving upon solutions.