Here is today's Pappas question of the day:
An equilateral triangle circumscribes a circle of radius sqrt(22/(3sqrt(3))). Find the area of triangle.
As I wrote earlier, I'll be switching to mostly algebra questions soon on my Pappas questions soon, as I'll be giving actual eighth grade systems of equations. But here's a geometry question to tide us over, since this is technically a geometry blog.
This question contains a common Pappas trick -- the apothem is given as sqrt(22/(3sqrt(3))), a very strange-looking number. But this is necessary to make the answer a whole number (the date). This happens often with equilateral triangle problems -- the area of an equilateral triangle with a integer side, radius, or apothem can never be an integer, but must contain a sqrt(3) factor. Even if the side, radius, or apothem itself contains a sqrt(3) factor, the area still has a sqrt(3) factor. In order to make the area a whole number, the side, radius, or apothem must contain the fourth root of 3.
When I see this sort of problem, it's actually easier just to use a variable, such as a for apothem, rather than the unwieldy sqrt(22/(3sqrt(3))). Then we substitute it back in later on.
The one thing I know about equilateral triangles is that they contain several 30-60-90 triangles. The actual proof of this is somewhat tricky -- the key is that angle bisectors of a triangle are concurrent -- they meet at the incenter of circle. Since the triangle is equilateral, it's equiangular -- the angle bisectors cut the 60 degree angles in to 30 degree halves. Along with the fact that angle bisectors of an isosceles (or equilateral) triangles are also altitudes, we conclude that the angle bisectors divide the whole triangle into six congruent 30-60-90 triangles. The apothem a is the short leg of these triangles, with the long leg a sqrt(3) and hypotenuse 2a. The original equilateral triangle has a base twice the length as the long leg, or 2a sqrt(3), and a height that of the short leg and hypotenuse combined, or 3a. So the area of the large triangle works out to be 3sqrt(3) a^2.
It only remains to plug in the given value of a:
a = sqrt(22/(3sqrt(3)))
a^2 = 22/3sqrt(3)
3sqrt(3) a^2 = 22
So the area is 22 -- and today's date is the 22nd.
Today is Day 104 here at the middle school, but it's only Day 100 in kindergarten. It's the first of three party days to occur at our K-8 school in the next few weeks.
Eighth grade, meanwhile, isn't anything like a party today. The students begin solving systems -- perhaps one of the most difficult topics in Common Core 8. On one hand, it's surprising that systems would even be included in the eighth grade standards -- but then again, I did say that every first semester Algebra I topic seems to appear in the Common Core 8 standards, and that would include systems of equations.
I remember back when I was student teaching an Algebra I class. We fell behind, and we had to rush through the systems of equations. We only had time to teach one method -- and it was graphing! A few years later, I told this story to an Algebra I teacher at another school, and she was surprised that we didn't choose elimination as our one method to teach, rather than graphing!
In some ways, I'm in the same predicament now. I'm rushing to cover all of the major content, or MC, standards this year, so today I only have time to teach one method of solving systems. This time, we haven't even reached graphing of lines yet, so instead of graphing, I choose substitution.
Why am I favoring substitution over elimination? Well, as it turns out, the Illinois State online homework that I'm required to give favors substitution. Indeed, the students are asked to solve four systems, and three of them already have x or y isolated in one of the equations. The fourth system is actually graphing -- but no equations are given. The lines are already graphed, and the students only need to identify the point of intersection. As it turns out, this is a lousy question -- the correct answer is given as (13, 2q), which is clearly a typo. I suspect that the q is supposed to be 9, and that the intended solution is (1.3, 2.9), but there are no decimals given in the computer solution.
Today's substitution lesson took a while to get through -- I suspect that part of it is that some students still struggle with multi-step equations. If you think about it, a substitution problem nearly always leads to a multi-step equation -- after all, if one equation is already solved for y, then we're to substitute an expression containing x into another equation also containing x. Students must know what to do next based on whether the two x-terms are on the same or opposite sides of the equation.
Of course, this means that equations not already solved for x or y are out of the question! We know that some systems are easy to solve with elimination (such as x + y = 4, x - y = 2) while others require multiplication before elimination can take place. Those students who struggle with multi-step equations might find elimination easier in some cases.
By the way, I often refer to elimination as "the method of three names." That's because I've seen three different names for this method -- the other two are "addition-subtraction" and "linear combinations."
I may introduce elimination as part of tomorrow's Learning Centers. This will be the first centers day with something on the line -- a major test is coming up on Friday. At least I've done centers once before, so hopefully they'll work out well tomorrow.
There isn't much of a science lesson today. It's the last Wednesday of the month, so there's already an Aspiration Assembly taking up much of the science period. (See my December 14th post for more info here.) Then afterward, the dean comes in to warn the students about gentrification yet again. (See my January 6th post for more info here.) There ends up being only 20 minutes for science. But, already anticipating less time for science, I have the students log on to Study Island and start answering questions related to their recent science projects. After all, in order to pass the California Science Test they'll need to be able to answer questions.
The sixth graders today are solving equations for the first time. But the equations given in the Illinois State text are trivial to solve -- the equations already have x or y isolated! The first equation they solve today is 5 + 7 = q. Yes -- there's that letter q as a variable again! One of my students sees q and thinks it's actually a 9 -- until I asked her whether 5 + 7 = 9. As usual, the seventh graders don't meet on Wednesdays.
This is a two-day post. My next post will be on Friday.
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