Area of a triangle: Equal cross-sections (Part 2)

Let’s take a second look at the familiar formula for the area of a triangle, $A = \displaystyle \frac{1}{2}bh$ .

The picture above shows three different triangles: one right, one obtuse, and one acute. The three triangles have bases of equal length and also have the same height. Therefore, even though the triangles have different shapes (i.e., they’re not congruent), they have the same height.

Let’s take a second look at these three triangles. In each triangle, I’ve drawn in three “cross-section” line segments which are parallel to the base. Notice that corresponding cross-sections have equal length. In other words, the red line segments have the same length, the light-blue line segments have the same length, and the purple line segments have the same length.

Why is this true? There are two ways of thinking about this (for the sake of brevity, I won’t write out the details).

Algebraically, the length of the cross-section increases linearly as they descend from the top vertex to the bottom base. This linear increase does not depend upon the shape of the triangle. Since the three triangles have bases of equal length, the cross-sections have to have the same length.
Geometrically, the length of the cross-sections can be found with similar triangles, comparing the big original triangle to the smaller triangle that has a cross-section as its base. Again, the scale factor between the similar triangles depends only on the height of the smaller triangle and not on the shape of the original triangle. So the cross-sections have to have the same length.

So, since the three triangles share the same height and base length, the three triangles have the same area, and the corresponding cross-sections have the same length.

The reverse principle is also true. This is called Cavalieri’s principle. From Wikipedia:

Suppose two regions in a plane are included between two parallel lines in that plane. If every line parallel to these two lines intersects both regions in line segments of equal length, then the two regions have equal areas.

In other words, if I have any kind of shape that has cross-sections that match those of the triangles above, then the shape has the same area as the triangles. Geometrically, we can think of each triangle a bunch of line segments joined together. So while the positioning of the line segments affects the shape of the region, the positioning does not affect the area of the region.

For example, here are three non-triangular regions whose cross-sections match those of the above triangles. The region on the right is especially complex since it has a curvy hole in the middle, so that the cross-sections shown are actually two distinct line segments. Nevertheless, we can say with confidence that, by Cavalieri’s principle, the area of each region matches those of the triangles above.

Though we wouldn’t expect geometry students to make this connection, Cavalieri’s principle may be viewed as a geometric version of integral calculus. In calculus, we teach that the area between the curves $x = f(y)$ and $x = F(y)$ is equal to

$A = \displaystyle \int_{y_1}^{y_2} [F(y) - f(y)] \, dy = \displaystyle \int_{y_1}^{y_2} d(y) \, dy$

where $d(y) = F(y) – f(y)$ is the difference in the two curves. In the above formula, I chose integration with respect to $y$ since the $y-$coordinates are constant in the above cross-sections. The difference $d(y)$ is precisely the length of the cross-sections. As with the triangles, the positioning of the cross-sections will affect the shape of the region, but the positioning of the cross-sections does not affect the length $d(y)$ and hence does not affect the area $A$ of the region.

Area of a triangle: Base and height (Part 1)

This begins a series of post concerning how the area of a triangle can be computed. This post concerns the formula that students most often remember:

$A = \displaystyle \frac{1}{2} b h$

Why is this formula true? Consider $\triangle ABC$ , and form the altitude from $B$ to line $AB$ . Suppose that the length of $AC$ is $b$ and that the altitude has length $h$ . Then one of three things could happen:

Case 1. The altitude intersects $AC$ at either $A$ or $C$ . Then $\triangle ABC$ is a right triangle, which is half of a rectangle. Since the area of a rectangle is $bh$ , the area of the triangle must be $\displaystyle \frac{1}{2} bh$ .

Knowing the area of a right triangle will be important for Cases 2 and 3, as we will act like a good MIT freshman and use this previous work.

Case 2. The altitude intersects $AC$ at a point $D$ between $A$ and $C$ . Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle ABD + \hbox{~Area of~} \triangle BCD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} b_1 h + \frac{1}{2} b_2 h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b_1 + b_2) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$

Case 3. The altitude intersects $AC$ at a point $D$ that is not in between $A$ and $C$ . Without loss of generality, suppose that $A$ is between $D$ and $C$ . Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle BCD - \hbox{~Area of~} \triangle ACD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t) h + \frac{1}{2} t h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t-t) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$

Engaging students: Completing the square

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Claire McMahon. Her topic, from Algebra: completing the square.

There were a lot of famous mathematicians that contributed to the notion of completing the square. The first of the mathematicians was that of the Babylonians. This culture started the notion of not only solving the quadratics but of arithmetic itself. The Babylonians started with the equations and then proceeded to solve them algebraically. Back then; they used pre-calculated tables to help them with solving for the roots. They were basically solving by the quadratic equation at this point. The man that came along has a very hard name to not only pronounce but to spell, and I will do my best. I will refer to him as Muhammad from here on out but his full name, or one of the common names to which he is referred is Muhammad ibn Musa al-Khwarizmi. He developed the term algorithm, which led to an algorithm for solving quadratic equations, namely completing the square.

The notion of completing the square has gone through a series of transformations throughout the history of mathematics. As mentioned before the Babylonians started with the notion and increased the knowledge by developing the quadratic formula to find the roots of a given quadratic equation. This spurred the thought that I can solve any equation and find its solution and roots by completing the square. Muhammad brought this notion to us, of which was mentioned before. More specifically the text that he developed was “The Compendious Book of Calculations by Completion and Balancing.” This book of course has been translated several times over but the general idea is laid out in the title. Modern mathematicians have developed a less compendious form that is now being taught in the math classes today. They take on many different forms and can be taught with manipulates as well.

The fabulous part to the story is there are a lot of resources that help the kids of today to deal with this “trick” of the math trade. There are numerous You Tube videos on the different methods of which show every step along the way with encouraging thoughts. Another great online resource is any of the math websites. I find it a little unfair that these resources were not readily available when I was struggling with such concepts. One of my personal favorites is the PurpleMath.com website. This website breaks everything down to basically that of a fourth grade level. They have pictures and fun problems to work out on your own. My favorite part is that you get your answers checked instantaneously to build the self-confidence and self-efficacy it takes to be a successful student. These particular websites are great tools for teachers as well, as they have a lot of great examples that can be used in the classroom and different ways that a student might present and calculate a problem.

A curious square root (Part 2)

There are two natural ways of computing $\sin 15^o$ using trig identities.

Method #1.

$\sin 15^o = \sin(45^o - 30^o)$

$\sin 15^o = \sin 45^o \cos 30^o - \cos 45^o \sin 30^o$

$\sin 15^o = \displaystyle \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

$\sin 15^o = \displaystyle \frac{\sqrt{6} - \sqrt{2}}{4}$

The same expression would be obtained if we had started with $15^o = 60^o - 45^o$ .

Method #2. Since $15^o$ is in the first quadrant,

$\sin 15^o = \sin \displaystyle \left( \frac{1}{2} \cdot 30^o \right)$

$\sin 15^o = \displaystyle \sqrt{ \frac{1 - \cos 30^o}{2} }$

$\sin 15^o = \displaystyle \sqrt{ \frac{1 - \displaystyle \frac{\sqrt{3}}{2}}{2} }$

$\sin 15^o = \displaystyle \sqrt{ \frac{2 -\sqrt{3}}{4} }$

$\sin 15^o = \displaystyle \frac{ \sqrt{2 -\sqrt{3}}}{2}$

Therefore,

$\displaystyle \frac{\sqrt{6} - \sqrt{2}}{4} = \displaystyle \frac{ \sqrt{2 -\sqrt{3}}}{2}$ ,

which may be verified by squaring both sides.

A curious square root (Part 1)

Here’s a square root that looks like a total mess:

$\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}}$

Just look at this monstrosity, which has a triply-embedded square root! But then look what happens when I plug into a calculator:

Hmmm. How is that possible?!?!

I’ll give the answer after the thought bubble, if you’d like to think about it before seeing the answer.

Let’s start from the premise that $\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}} = 3$ and work backwards. This isn’t the best of logic — since we’re assuming the thing that we’re trying to prove in the first place — but it’s a helpful exercise to see exactly how this happened.

$\sqrt{5 - \sqrt{6} + \sqrt{22+8\sqrt{6}}} = 3$

$5 - \sqrt{6} + \sqrt{22+8\sqrt{6}} = 9$

$\sqrt{22+8\sqrt{6}} = 4 + \sqrt{6}$

$22 + 8 \sqrt{6} = (4 + \sqrt{6})^2$

This last line is correct, using the formula $(a+b)^2 = a^2 + 2ab + b^2$ . So, running the logic from bottom to top (and keeping in mind that all of the terms are positive), we obtain the top equation.

This suggests a general method for constructing such hairy square roots. To begin, start with any similar expression, such as

$(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3$

$(2 - \sqrt{3})^2 = 7 - 4\sqrt{3}$

Then we create a nested square root:

$2 - \sqrt{3} = \sqrt{7 - 4\sqrt{3}}$

Then I get rid of the square root on the left hand side:

$2 = \sqrt{3} + \sqrt{7 - 4\sqrt{3}}$

Then I add or subtract something so that the left-hand side becomes a perfect square.

$25 = 23 + \sqrt{3} + \sqrt{7 - 4 \sqrt{3}}$

Finally, I take the square root of both sides:

$5 = \sqrt{23 + \sqrt{3} + \sqrt{7 - 4 \sqrt{3}}}$

Then I show the right-hand side to my students, ask them to plug into their calculators, and ask them to figure out what happened.

Engaging students: Measures of the angles in a triangle add to 180 degrees

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Claire McMahon. Her topic, from Geometry: the proof that the measures of the angles in a triangle add to $180^o$ .

One of the hardest concepts in math is learning how to prove something that is already considered to be correct. One of the more difficult concepts to teach could also be said on how to prove things that you had already believed and accepted in the first place. One of these concepts happens to be that a triangle’s angles are always going to add up to 180 degrees. Here is one of the proofs that I found that is absolutely simplistic and most kids will agree with you on it:

This particular proof is from the website http://www.mathisfun.com. This is a great website to simply explain most math concepts and give exercises to practice those math facts. For the more skeptical student, you can use a form of Euclidean and modern fact base to prove this more in depth. I found this proof on http://www.apronus.com/geometry/triangle.htm. Here you will see that there is no question as to why the proof above works and how it doesn’t work when you do a proof by contradiction.

I stumbled across this awesome website that very simply put into context how easy it would be to prove that a triangle’s angles will always add up to 180 degrees. In this activity you take the same triangle 3 times and then have them place all three of the angles on a straight line. This proves that the angles in a triangle will always equal 180 degrees, which is a concept that should have already been taught as a straight line having an “angle” measure of 180. The website for this can be found here: http://www.regentsprep.org/Regents/math/geometry/GP5/TRTri.htm.

The triangle is the basis for a lot of math. There is one very important person that really started playing with the idea of a triangle and how 3 straight lines that close to form a figure has a certain amount of properties and similarities to parallels and other figures like it. We base a whole unit on special right triangles in geometry in high school and never know exactly where the term right angle is derived. This man that made the right angle so important in math is none other than Euclid himself. While Euclid never introduced angle measures, he made it very apparent that 2 right angles are always going to be equal to the interior angles of a triangle. Not only did Euclid prove this but he did so in a way that relates to all types of triangles and their similar counterparts using only a straight edge and a compass, pretty impressive!!

Interdisciplinary studies (Part 4)

Shamelessly stolen from a friend:

How do you tell the difference between a plumber and a chemist?

Ask them to pronounce unionized.

Continued fractions and pi

I suggest the following activity for bright middle-school students who think that they know everything that there is to know about fractions.

The approximation to $\pi$ that is most commonly taught to students is $\displaystyle \frac{22}{7}$ . As I’ll discuss, this is the closest rational number to $\pi$ using a denominator less than $100$ . However, it is possible to obtain closer rational approximations to $\pi$ using larger numbers. Indeed, the ancient Chinese mathematicians were superior to the ancient Greeks in this regard, as they developed the approximation

$\pi \approx \displaystyle \frac{355}{133}$

It turns out that this is the best rational approximation to $\pi$ using a denominator less than $16,000$ . In other words, $\displaystyle \frac{355}{133}$ is the best approximation to $\pi$ using a reasonably simple rational number.

Step 1. To begin, let’s find $\pi$ with a calculator. Then let’s now subtract $3$ and then find the inverse.

This calculation has shown that

$\pi = \displaystyle 3 + \frac{1}{7.0625133\dots}$

If we ignore the $0.0625133$ , we obtain the usual approximation

$\pi \approx \displaystyle 3 + \frac{1}{7} = \frac{22}{7}$

Step 2. However, there’s no reason to stop with one reciprocal, and this might give us some even better approximations. Let’s subtract $7$ from the current denominator and find the reciprocal of the difference.

At this point, we have shown that

$\pi = \displaystyle 3 + \frac{1}{7 + \displaystyle\frac{1}{15.9965944\dots}}$

If we round the final denominator down to $15$ , we obtain the approximation

$\pi \approx \displaystyle 3 + \frac{1}{7 + \displaystyle\frac{1}{15}}$

$\pi \approx \displaystyle 3 + \frac{1}{~~~\displaystyle \frac{106}{15}~~~}$

$\pi \approx \displaystyle 3 + \frac{15}{106}$

$\pi \approx \displaystyle \frac{333}{106}$

Step 3. Continuing with the next denominator, we subtract $15$ and take the reciprocal again.

At this point, we have shown that

$\pi = \displaystyle 3 + \frac{1}{7 + \displaystyle\frac{1}{15 + \displaystyle \frac{1}{1.00341723\dots}}}$

If we round the final denominator down to $1$ , we obtain the approximation

$\pi \approx \displaystyle 3 + \frac{1}{7 + \displaystyle\frac{1}{16}}$

$\pi \approx \displaystyle 3 + \frac{1}{~~~\displaystyle \frac{113}{16}~~~}$

$\pi \approx \displaystyle 3 + \frac{16}{113}$

$\pi \approx \displaystyle \frac{355}{113}$

Step 4. Let me show one more step.

At this point, we have shown that

$\pi = \displaystyle 3 + \frac{1}{7 + \displaystyle\frac{1}{15 + \displaystyle \frac{1}{1 + \displaystyle \frac{1}{292.634598\dots}}}}$

If we round the final denominator down to $292$ , we (eventually) obtain the approximation

$\pi \approx \displaystyle \frac{52163}{16604}$

The calculations above are the initial steps in finding the continued fraction representation of $\pi$ . A full treatment of continued fractions is well outside the scope of a single blog post. Instead, I’ll refer the interested reader to the good write-ups at MathWorld (http://mathworld.wolfram.com/ContinuedFraction.html) and Wikipedia (http://en.wikipedia.org/wiki/Continued_fraction) as well as the references therein.

But I would like to point out one important property of the convergents that we found above, which were

$\displaystyle \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, ~ \hbox{and} ~ \frac{52163}{16604}$

All of these fractions are pretty close to $\pi$ , as shown below. (The first decimal below is the result for $22/7$ .)

In fact, these are the first terms in a sequence of best possible rational approximations to $\pi$ up to the given denominator. In other words:

$\displaystyle \frac{22}{7}$ is the best rational approximation to $\pi$ using a denominator less than $106$. In other words, no integer over $8$ will be any closer to $\pi$ than $\displaystyle \frac{22}{7}$ . No integer over $9$ will be any closer to $\pi$ than $\displaystyle \frac{22}{7}$ . And so on, all the way up to denominators of $105$ . Small wonder that we usually teach children the approximation $\pi \approx \displaystyle \frac{22}{7}$ .
Once we reach $106$ , the fraction $\displaystyle \frac{323}{106}$ is the best rational approximation to $\pi$ using a denominator less than $113$ .
Then $\displaystyle \frac{355}{113}$ is the best rational approximation to $\pi$ using a denominator less than $16604$ .

As noted above, the ancient Chinese mathematicians were superior to the ancient Greeks in this regard, as they were able to develop the approximation $\pi \approx \displaystyle \frac{355}{113}$ . For example, Archimedes was able to establish that

$3\frac{10}{71} < \pi < 3\frac{1}{7}$

Engaging students: The field axioms

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Maranda Edmonson. Her topic, from Pre-Algebra: the field axioms of arithmetic (the distributive law, the commutativity and associativity of addition and multiplication, etc.).

B. Curriculum: How can this topic be used in your students’ future courses in mathematics or science?

It is safe to say that the field axioms are used in all mathematics classes once they are introduced. As students, we know them to be rules for how to simplify or expand expressions, solving equations, or just manipulating numbers and expressions. As instructors, we know them to be a solid foundation for further mathematical understanding. “In mathematics or logic, [an axiom is] an unprovable rule or first principle accepted to be true because it is self-evident or particularly useful” (Merriam-Webster.com). Is the distributive property not useful? Isn’t the associative property self-evident? We learn these axioms, master them during the first lesson we encounter them, and they stick with us. Why? Because they are obvious “rules” that we use and apply to all aspects of mathematics. They are a foundation on which we, as instructors, wish to build upon a greater mathematical understanding.

B. Curriculum: How does this idea extend what your students should have learned in previous courses?

When students first begin to learn addition they are learning the associative property as well. Think about it – when kids learn about the expanded form of a number, they are already seeing that when you add more than two numbers together they equal the same thing, no matter what order they are being added in. For example:

$1,458 = 1,000 + 400 + 50 + 8 = (1,000 + 400) + (50 + 8) = (1,000 + 50) + (400 + 8)$

and so on. Kids tend to add numbers in the order that they are given. However, when they start learning little tricks (say, their tens facts), then they will start seeing how the numbers work together. For example: $3 + 4 + 7$ soon becomes $(3 + 7) + 4$ . Then, when students get into higher grades and begin learning multiplication, the commutative property becomes a real focus. When they are learning their multiplication facts, students are faced with $5 \times 7$ one minute, then $7 \times 5$ the next. They start seeing that it does not matter what order the numbers are in, but that when two numbers are being multiplied together, they will equal the same product each time.

E. Technology: How can technology be used to effectively engage students with this topic?

Math and music are always a good combination. Honestly, who doesn’t hum “Pop! Goes the Weasel” every time they need to use the quadratic formula? This YouTube video (the link is below) is of some students singing a song about the associative, commutative and distributive properties. The video is difficult to hear unless you turn the volume up, and the quality is not the greatest. However, the students in the video get the point across about what the axioms are and that they only apply to addition and multiplication. Note that you only need to watch the first three minutes of the video. The last minute and a half or so is irrelevant to the axioms themselves.

Full lesson plan: Designing a model solar system

Over the summer, I occasionally teach a small summer math class for my daughter and her friends around my dining room table. Mostly to preserve the memory for future years… and to provide a resource to my friends who wonder what their children are learning… I’ll write up the best of these lesson plans in full detail.

This was a fun activity that took a couple of hours: designing a model Solar System. I chose the scale so that most of the planets would fit on a straight section of sidewalk near my house; of course, the scale could be changed to fit the available space.

For my particular audience of students, I also worked through the basics of the metric system as well as decimals.

This lesson plan is written in a 5E format — engage, explore, explain, elaborate, evaluate — which promotes inquiry-based learning and fosters student engagement.

Model Solar System Handout

Model Solar System Lesson

Post Assessment

P.S. For what it’s worth, the world’s largest model solar system can be found in Sweden.