Honey Combs, Cell Towers, and Geometry Woes OH My!
by Cassidy Cash
Several students have contacted us in the last week or so
asking about their geometry course problems. Specifically they've been asking
about the real world example problems.
There are problems in Chapter 14 that ask students to
consider that the map for cell tower and cell phone coverage is a series of
hexagon shapes laid together and forming a honeycomb look. The book makes this
statement and then goes on to ask a series of questions about hexagons and
geometric propery application with each problem building on the last problem
until you arrive finally at a rather complex conclusion.
An activity you might consider to spice up these problems: Make time to allow students to look up the real cell towers, visit a library to find schematics, or talk with an expert. You might visit a cell phone company's website and explore what their "coverage area" really means. Have students combine the information into a report and give points for being able to include not only the concepts they are learning from Chapter 14, but perhaps a demonstrative "honey comb" drawing as well.
When they arrived at these Chapter 14 problems, my homework help students flooded me with emails.
"Mrs. Cash! Mrs.
Cash! I don't get it!! I don't know anything about cell towers!!!"
"Take Heart!", I told them, "Learning how to contruct cell towers is not the point here. Breathe. ok. Now let me show you what you are supposed to be learning. "
Here is an excerpt of our conversation(s).
Hey there!
Thanks for contacting
us! You've asked really good questions, let me see if I can help.
The answer to problem 39 is based
on the 30-60 Right Triangle Theorem (Theorem 51) on page 442 of this same
lesson. It says that in a 30-60 right triangle, the hypotenuse is twice the
shorter leg...
From problem 38, we know that BD = AB. So if AB = x, BD = x.
Triangle BCD is a 30-60 right triangle with hypotenuse BD and shorter leg BC,
so by the 30-60 Right Triangle Theorem, BD = 2BC.
So x = 2BC and BC = x/2.
It is also true that BC = AC - x, but the problem said to express BC in terms
of x, not something else and x.
Does that help? Contact me
if I can clarify anything, or if you come across more questions.
God Bless,
Cassidy Cash
And their response?
Thank you so much, Mrs. Cash! I
understand now.
Man, we
love that answer. It is so awesome to be able to help students. It's not only amazing, but at Homework Help, it's our job. The
students we work with consistently do better in their work and raise their math grades at home and on the ACT/SAT.
I love my job.
~Cassidy Cash is the Executive Producer for AskDrCallahan, she contributes to the company through the Algebra 1 course, Homework Help, and much more. We thank her for contributing this article.
