Monday, April 9, 2018

Activity: Hexagons (Day 140)

Today on her Mathematics Calendar 2018, Theoni Pappas writes:

The area of this regular hexagon is (121 1/2)sqrt(3) sq. units. What is the length of each side?

There are two ways to solve this problem -- the "Second Edition" and "Third Edition" methods. The names refer to the editions of the U of Chicago text. In neither edition are we given the area of a regular hexagon and asked to find the side -- instead the side is given and the area is to be found. So for both methods, we will find the area of a regular hexagon of side s. Then we'll set the area equal to the given value (121 1/2)sqrt(3) and solve for s.

Actually, our Second Edition of the text doesn't discuss the area of the hexagon at all. Instead, Lesson 14-1 only gives an example asking for the height:

Example 2 (Second Edition, Lesson 14-1):
A zoologist measured the length of the side of a cell in a natural beehive (a tessellation of regular hexagons) as 8.0 mm but forgot to measure the height h of each cell. Approximate the height to the nearest 0.1 mm.

Solution:
The figure below isolates one cell. DH is the desired height. Angle DIH = 120 and DI = IH from the properties of a regular hexagon, so Triangle DIJ is a 30-60-90 right triangle with hypotenuse 8.0. Let IJ, the shortest side, be x. Then, using the 30-60-90 Triangle Theorem, the hypotenuse DI = 2x = 8.0, so x = 4.0. Longer leg DJ = x sqrt(3) = 4sqrt(3). So:

DH = 2 * 4sqrt(3) = 8sqrt(3) = 13.9 mm.

OK, so now we'll redo this problem so that the side length is s rather than 8:

Solution (with side length s):
The figure below isolates one cell. DH is the desired height. Angle DIH = 120 and DI = IH from the properties of a regular hexagon, so Triangle DIJ is a 30-60-90 right triangle with hypotenuse s. Let IJ, the shortest side, be x. Then, using the 30-60-90 Triangle Theorem, the hypotenuse DI = 2x = s, so x = s/2. Longer leg DJ = x sqrt(3) = (s/2)sqrt(3). So:

DH = 2 * (s/2)sqrt(3) = s sqrt(3).

OK, but notice that the Second Edition doesn't give the area of the hexagon at all. But we notice that it's possible to draw in a second altitude so that the two altitudes divide the hexagon into a rectangle and two obtuse triangles. The length of the rectangle equals the height of the hexagon (s sqrt(3)) and the width of the rectangle equals the side of the hexagon (s). Thus its area is:

Area of rectangle = s sqrt(3) * s = s^2 sqrt(3).

As for the triangles, the base of each triangle equals the height of the hexagon (s sqrt(3)) and the height of each triangle equals the shorter leg of the right triangle we found earlier (x = s/2). Thus:

Area of each obtuse triangle = 1/2 * s sqrt(3) * s/2 = (s^2)/4 sqrt(3).

Therefore the area of the hexagon is:

Area of hexagon = Area of rectangle + Area of 2 obtuse triangles
                            = s^2 sqrt(3) + (2s^2)/4 sqrt(3)
                            = (3s^2)/2 sqrt(3).

Meanwhile, the Third Edition method of finding the area involves the apothem formula -- a formula not mentioned in the Second Edition:

Example 3 (Third Edition, Lesson 8-7):
Find the area of a regular hexagon with side length 10.

Solution:
Draw regular hexagon HIJKLM with apothem OP.

The perimeter of the hexagon is 6(10) = 60. To find the apothem, notice that KP = 5 because it is half of the side length. Also notice that Angle KOJ = 360/6 = 60, so Angle POK = 30. Thus, Triangle KPO is a 30-60-90 triangle in which 5 is the length of the shorter leg. So, OP = 5sqrt(3).

Area(HIJKLM) = 1/2 ap
                          = 1/2(5sqrt(3))(60)
                          = 150sqrt(3).

Solution (with side length s):
The perimeter of the hexagon is 6s. To find the apothem, notice that KP = s/2 because it is half of the side length. Also notice that Angle KOJ = 360/6 = 60, so Angle POK = 30. Thus, Triangle KPO is a 30-60-90 triangle in which s/2 is the length of the shorter leg. So, OP = s/2 sqrt(3).

Area(HIJKLM) = 1/2 ap
                          = 1/2(s/2 sqrt(3))(6s)
                          = (3s^2)/2 sqrt(3).

In either case, we obtain the same area, (3s^2)/2 sqrt(3). To solve the Pappas problem, we finally set this equal to the area that she gives us:

(3s^2)/2 sqrt(3) = 121 1/2 sqrt(3)
(3s^2)/2 = 121 1/2
3s^2 = 243
s^2 = 81
s = 9

Therefore the desired side length is 9 -- and of course, today's date is the ninth.

Our students can't use the apothem formula since our old edition of the U of Chicago text doesn't define the apothem. And they can't use the other method because it depends on Chapter 14 material, which we haven't quite reached yet -- even though we're oh so close.

In fact, recall that today is supposed to be the Chapter 13 Test, but we gave the test last Friday in order to avoid a Monday exam. At any rate, Chapter 14 is supposed to start on Day 141, which will be tomorrow. So today is an activity day instead.

Hey -- I just realized something. We spent so much time working on a Chapter 14 question the day before we begin that chapter. Many teachers give opening activities on the day before they properly begin a unit.

So why don't we make today's Pappas question into an opening activity for Chapter 14? Originally, I was going to post a closing activity for Chapter 13 -- and in fact, I still will, since it's a Logo activity from last year.

This is what I wrote last year about today's Logo activity:

I decided to take Logo a step further and actually include an activity page, based on Exploration Question #18 from the text. Maybe it's because of all that computer time I had in my classroom, even though that's IXL and Study Island, not Logo!

And here again is the Logo link I provided in last Thursday's post:

http://www.calormen.com/jslogo/

Of course, not every teacher will want to give the students a Logo activity. So let me give you teachers a choice -- you can give either the Logo activity or a hexagon area problem based on today's Pappas question. I like the idea of having the students draw a regular hexagon in Logo, so that both choices lead to regular hexagons -- either draw one in Logo or find the area of one.

to hexagon :length
  repeat 6 [fd :length rt 60]
end
clearscreen
hexagon 100

As for the hexagon area problem, I decided not to give the Pappas problem exactly as written. Instead I'll make two changes.

First of all, let's make the side length an even number. The presence of 2 in the denominator of the area formula forces us to take half of an odd number, which explains why the area of a hexagon with side length 9 is 121 1/2 sqrt(3). Notice that the examples in the two editions of the U of Chicago text have side lengths 8 and 10 instead of 9.

The second change is to give the side length and ask for the area, not vice versa. It's clear why Pappas asks for the side rather than the area -- the area formula contains a factor of sqrt(3). Obviously, the answer can't work out to equal the date if it contains sqrt(3).

Suppose Pappas wanted to ask for the area of a regular hexagon -- is it possible to make the answer equal a valid date? Unfortunately, making the side length a multiple of sqrt(3) doesn't solve the problem, since the formula involves the side length squared. For example:

s = 2sqrt(3)
A = (3s^2)/2 sqrt(3)
A = (3(2sqrt(3))^2)/2 sqrt(3)
A = (3(12))/2 sqrt(3)
A = 18 sqrt(3)

In order to make sqrt(3) vanish, the side length must be a multiple of the fourth root of 3. We try it:

s = 2sqrt(sqrt(3))
A = (3s^2)/2 sqrt(3)
A = (3(2sqrt(sqrt(3)))^2)/2 sqrt(3)
A = (3(4sqrt(3))/2 sqrt(3)
A = 6sqrt(3)sqrt(3)
A = 18

If Pappas wanted to, she could give the above problem on the 18th of the month. If I recall correctly, she did give either this or a similar problem a few years ago.

Our goal today, of course, is not to scare students away with fractions or the fourth root of 3. So we keep things simple by making the side length a simple even number. Once we do so, the students will reach the main challenge of the problem -- finding the area of the hexagon without already knowing the 30-60-90 Triangle Theorem.

I suspect that the method of the Third Edition is easier, even if the students have never heard of the apothem formula. Hopefully, the students will think to divide the hexagon into six triangles -- so all we have to do is multiply the area of one of them by six. The base of each triangle equals the length of the side, so all that's needed is to find the height.

The key is to show that all six small triangles are equilateral -- once this is known, then they can divide each equilateral triangle into two right triangles and use the Pythagorean Theorem. They'll have to do it the same way the text does it -- by finding relevant angle measures.

The fact that the six smaller triangles are equilateral also explains why the construction for a regular hexagon works. The side of the hexagons equals the radius of the circumscribed circle. So this also makes a great excuse to return to that construction -- which, as you may recall, is part of the Common Core Standards yet doesn't appear in either edition of the U of Chicago text.

Let's use the rest of this post to resolve a few loose ends. First of all, Sarah Carter of Oklahoma finally acknowledges the teacher strike in her state:

https://mathequalslove.blogspot.com/2018/04/five-things-volume-15.html

She writes that school was closed last week for the strike, but even though the issues haven't been resolved, her district is now open. (Last week was not spring break in her district -- indeed, her spring break was the week before vacation week on the blog calendar. Most likely, spring break divides the quarters in her district -- as opposed to my district, where break is a few days after the start of the fourth quarter.)

In her weekly "Monday Must Reads" series, Carter links to two classes calling themselves "Pre-AP" (well, I assume PAP means "Pre-AP"). One of them had a STEM project to learn about direct variation, while the other created art out of parabolas. You'd better hope that SteveH doesn't find out about these projects -- he already opposes the idea of Pre-AP in the first place, and of course he'd argue that STEM and art projects do not prepare students to earn a 5 on the AP.

Oh, and speaking of SteveH, the traditionalist debate at Barry Garelick's website is heating up. The Ed Week article I mentioned in my last post generated another Garelick post and several SteveH comments in response. I don't wish to label this yet another "traditionalists" post, so let me just give you the gist of it. Garelick is still upset that so many people think that "understanding" is more important than knowing how to solve basic math problems. SteveH criticizes something called "Polya's 4-step process of problem solving." Traditionalists say that math should be about solving problems, not a mysterious process called "problem solving" or how to "problem solve."





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