Challenge Problem - May 22, 2014

Hello All!

I have a new challenge problem for us to first figure out here and then take to our classrooms and report back how it went.  This one is definitely a mathematical reasoning problem, so let's put on our thinking caps and discuss the solution, how you arrived at the solution and what standards or mathematical practices do you see at work.  Here is the problem:

The area, A, in square inches, of a parallelogram with a height of 4 inches is given by the equation  A = 4L, where L is the length, in inches, of the base of the parallelogram.  The table shows the area, M, in square inches, of a triangle with a height of 4 inches and a base of p inches.

p (inches) M (square inches)
1 2
3 6
36 72
52 104

The area of the triangle is ___________ the area of the parallelogram.

Fill in the blank.

Happy Discussing - I will post the solution next week!

Brooke Istas



I believe it is half.  When I taught this before I used paper cut outs of squares, rectangles, and had the class cut from one corner to the far corner and see if the shapes were the same size.


Thank you for your post!  I love the idea of using math manipulatives in the classroom but I hadn't considered it for this problem.  What a great way for the learners to come up with their own understanding using hands-on tools!  Do you feel that the manipulatives enhanced the learning?


The area of a triangle with base b will have half the area of a parallelogram with base b ... but in this question they don't specify that the bases are the same.    Teacher I worked with used cheezits, which are one square inch :)   I love the idea of working with paper cutouts, though.   

I also carry this over to circles and show how when we say "r squared" it's really to have the radius going horizontally and the radius going vertically -- like "base times height" -- and then... four would be too big, but pi is *just right* for getting the rest of the circle's space. 

Okay, this problem has me rather confounded. 

The area of our parallelogram is "4L" -- so, since 4L = HL  (area of a parallelogram is base times height), the area of our parallelogram is 16 and its base and height are four.

However, what does that have to do with the triangles, which all have different variables?   

What is "the triangle"?   I see a table of lots of different triangles.    


Is the point of this  to try to construct a problem from utterly ambiguous information?   So we can understand what it feels like to be a student who doesn't understand math?   


First, thank you for your response.  I appreciate you taking the time to try this problem and then posting your thoughts about it.  It is good to know that I am not the only one who first thought this problem was a bit confusing with all the information given.  

This problem is very similar to the problems our learners are seeing on the *NEW* GED Mathematical Reasoning Test.  The purpose of me posting it here is so that the community can discuss the answer, how they arrived at their solution, and also to utilize this problem in their classroom and report back on what happened.  It is not meant to make you feel like a student who does not understand math.

The new Pearson Vue Mathematical Reasoning test is different from the old GED test in that is requires student to interpret the information given to arrive at a solution.  Over the next few weeks, I will be posting problems that are similar to what our students have been seeing.  My hope is that the community will begin to discuss these problems so that instructors begin to feel comfortable presenting these types of questions and develop a deeper understanding of mathematical concepts.

I am going to hold off explaining the solution until Friday because I would like some other members of the community to speak up (as we know from being in the math classroom, once the answer is given the discussion stops).  So, colleagues in the field…let’s talk about this problem.

Thanks again, Sue for your post!

P.S.  I found that drawing pictures helped me.


I'm pretty sure I understand the problem.   

I'm just finding it perverse that students seem to be expected to interpret "the triangle" not as a specific triangle but as triangles in general, if indeed the answer is "half." 

I'm also finding it totally frustrating not to be able to refer to the original post while responding, so I'm just going to stop right now.  

Now, I could do some contrived algebra and say that since the area of all triangles is 1/2 bh and  our "triangle" even though it's really a lot of triangles  has a height of 4, therefore has an area of 2p.   Why they don't let the base be "b" is also silly. 

I could then plug that into the question's wording 

Area of triangle = x times (that's what of means) area of parallelogram. 

So, 2p = x times 4L 

x = 2times the base of the triangle, divided by the product of four and the  length of the square.   

x = 1/2(p/L)   or p/(2L)      So "the area of the triangle is p/2L of the area of the parallelogram" -- not language that humans or math books use.   

Another weirdness is that students are not expected to know that the area of a parallelogram is the base times the height, or that the area of a triangle is half the base times the height -- both pretty basic concepts (even if so many students Only Know THe Formula).  




By looking at the table, I can see that the area of the triangle is twice the base (M=2p).

By using the formula, I can see that the area of the parallelogram is four times the base (A=4L). This might be all the student needs to observe.

If I use some of the values for p and insert them as L in the parallelogram formula, I can create a third column of the table that shows the area of the parallelogram using the same base length. That column would start: 4, 12, 144, 208 ... By examining these two columns, the student can see that the area of the triangle is half the area of the parallelogram.

Of course, my hope is that the student would already know this and would simply be looking for verification. The problem would be much more straightforward if the variable for the base (p in the table and L in the formula) was consistent throughout the problem. I agree that b would be a more logical choice. (This is especially true since p, if it stands for anything, seems like it ought to stand for parallelogram--or possibly perimeter.)


Can you tell me what that's based on?   I keep looking but not finding where it says that the bases of the two shapes are equal.   

Yes.   Can you explain why I should conclude that if the heights are the same, the bases are also the same, except for the fact that if they are the same, the problem gets a numerical answer? 

Mathematical reasoning -- they are different things and were assigned different variables -- would say otherwise.

(And this week's problem *does* have multiple correct answers, right?) 

... and part of me is, of course, asking "Why is that Sue JOnes obsessing over that?"  -- but these are tests with rather high stakes, and the student who is *good* at math reasoning will know that 1/2 is not a correct answer.  If 1/2 is the "answer,"  The problem *breaks* the reasoning practices it's supposed to be testing (unless I'm missing something, which is always possible).  P is the base of one thing, L is the base of a different shape.   Just because they are both bases doesn't mean they are both equal... 

    This reminds me of the assorted viral examples of some horrific "problems" in some of the math materials being associated with Common Core. 


I want to help you understand this probably as much as you want to.  So, let me try another way, it seems that you are concerned about the bases being different.  The problem states that the heights are the same (not directly) but it does say the height of the parallelogram is 4 and the height of the triangle is 4.  With this in mind, and let's use formulas with letters we are familiar with...

Parallelogram is A = bh

Triangle is A = 1/2 bh

The commutative property of multiplication means that the order you multiply these numbers does matter, right?  So, you could think that the bases are the same if you want.  Is this helping you understand the solution better?



Rachel and all,

Thank you for your response.

I agree that the variables not being labeled what we are used to (b for base and etc.) is a bit uncomfortable.  But, as we work more and more with variables what they are labeled is not nearly as important as understanding that the letter (whatever it may be) represents something.  I find that my learners struggle with the idea of letters in formulas, equations, graphs, etc.  Am I alone in this?  Is anyone else having the same experience?


My students do struggle with the concept of variables standing for things, as well as what "equality" is (and what I've read says it's pervasive - see  ).   

I think it's totally valid to test this by using different letters to assess whether students recognize the "area = base times height" for a parallelogram even if we use different letters.

To help students understand this, I've started walking through that conceptual process far more often.   My programming course shone a blazing spotlight on the reality that I sometiems think "hey! I showed you, you understood it, and now you can apply it, right?"  D'oh.   I can understand something perfectly in the context of instruction ... but still need to walk through it in *my* brain -- even in the same context -- a few times... and then in different contexts. 

So... I spend some time plugging in numbers to show students that, yea verily, 2x + 3x = 5x -- >  no matter what x is!   And that x + x is 2x not x^2...

... and then that (x + y)^2 is not x^2 + y^2...

Equality is another thign that I've started talking about conceptually every chance I get.  No, the equals sign isn't "what comes before the answer."   It's more like a balance... 

The difference in approach for me was getting away from "help this student solve this math problem" to "hey! every math problem is a golden opportunity to review important concepts!"   and... so far, it *hasn't* been onerously time-consuming.   

Hello All!

Just a reminder that I will be posting the solution to the challenge question tomorrow AND there will be a new challenge question to think about and discuss next week.  So, if you haven't chimed in with your response - now is the time to do so!  

Happy Thinking!


The solution to this math problem is, 

The area of a triangle is half the area of a parallelogram

How I arrived at my solution:


Area of a Parallelogram is A=4L, where 4 is the height (or width) in inches.  

Several areas and bases of triangles who all have the same height of 4 inches

Where I started:

I started with the table, even though the formula for Area of a Triangle isn't given I thought I should try to see if a pattern existed.  Just given M = 4p, is a true statement.  This is what I found,

2 = 4 (1)  <---doesn't equate but appears to be half of the right side

6 = 4 (3) <---seems to be the same as above, doesn't equate but appears to be half of the right side

36 = 4 (72) <----final check, we need to add 1/2 to the given formula of M = 4p for this table to be correct.

So I constructed an argument that looked for a pattern - then I changed the formula from just base and height to M = 1/2 (4) p or M = 2p (formula for triangles in this problem).

Thus, comparing that given formula for parallelograms A = 4L  and the formula I constructed for triangles, I concluded that the area of the triangle is half the area of a parallelogram.

As you can tell this problem was not like the straight forward problems that used to be on the old version of the GED exam.  That's the purpose of these weekly challenge problems is so the community (Math and Numeracy) can be a support area, a safe place, to discuss these problems and about how we arrived at the solutions.  Please do not continue to be a lurker, be a contributor to the community!  Let's help each other understand these new math requirements better!


For your conclusion to be true, 

the bases must be equal. 

When you can show me where in that problem it indicates that the bases are equal then I'll agree with that conclusion.   

My math teacher associates agreed that the solution to the problem would have to have two variables to be true. 

With valid mathematical reasoning, we cannot replace one variable with another.  THe base of the triangle is one thing and the base of the parallelogram is another. 

THey were given different letters.

According to the *mathematical reasoning* this question is supposed to be assessing, therefore, they are independent of each other.   If I say the height of my sister is S and my brother's height is B... and I want to calculate their Body Mass Indices... I can't swap variables. 

THis will be "fine" if the test taker makes the same false assumption as the test maker made... which I woudl imagine would happen fairly often.  However, I do apologize for being a misfit about this, but I have this really strange idea that we shouldn't say something is "correct" when it is false. 

If I'm keeping anybody else from participating... I do apologize and I'll cease and desist and take up lurkdom for a while... and I have an app to work on... but if somebody wants me to try to break it down further to explain, I'd be willing.   


I do not want you to feel that you should cease and desist this conversation but I do want other community members to help me try to explain this problem to you. Furthermore, I realize that you are not attacking me about this problem or my reasoning.  As I have stated before, this community should be a safe environment for practitioners to speak their thought processes without being made to feel attacked.  I also agree that we should be careful stating that things are correct when it is false; but conversely, I want to caution us in saying that something is false prematurely, too.

I agree with your argument about Body Mass but in this problem we aren't swapping bases.  There is a common measurement of 4 inches.  Both shapes are 4 inches in measurement.

A = 4L but the formula for the other is not given.  Just a table of values.  The only consistent is the 4 inches in height.

Therefore, if you don't know the formula for a triangle you could derive it, as I did in the solution that I posted.  The table has a pattern that fits the formula for area of a triangle, which of course is, A = 1/2 bh (I am going to use b since p and M seems to be an issue) so let's put the height we are given into this equation. A = 1/2 b 4 (without simplifying).

The problem doesn't state that L and p are equal, but to compare areas need to use some of the same values to see what is going on with the Areas using the same heights. 

Parallelogram Areas, A = 4L                                                                                                                          Triangle Areas, A = 1/2 b h

Area Height Length 4 4 1 8 4 2 16 4 4 Area Height Base * 1/2 2 4 1 *1/2 4 4 2 * 1/2 8 4 4 * 1/2





From the graph, the Area of the Triangle is 1/2 that of the Parallelogram.

Does this help?  If not, let us know.  Also, please other community members, see if you can explain this problem differently.  Remember, this is a safe environment for a healthy debate.

Brooke   has a neat diagram of this problem. is a really popular site amongst my math tweeps ;) 

I completely understand that triangles have half the area of any parallelogram w/ same base and height. I teach this a few dozen times a year, with graphs and pictures... but that's not what that problem asked.  


Sue is right that the problem never specifies which triangle is being compared to which parallelogram. In the final fill-in-the blank section, they should have said something along the lines of, "The area of each triangle is ___________ times the area of the parallelogram with the same length base." If they had stipulated at some point that the base lengths need to be the same, then it wouldn't matter that the variables are different, since they're essentially telling you to set p equal to L. Still, it's a clunky problem. They are, as the Car Talk guys say, "obfuscating" the basic rule they are trying to use, and they may have over-obfuscated a bit in their zeal to get a harder problem.

When I run into situations like this in class, I praise the student(s) who points out the issue and make sure everyone understands why the difference in wording is important if possible by having them do both what the problems says and what the writer probably meant. I know that because I see countless problems of this ilk every week, I get into the habit of seeing what I assume will be there. This is a serious problem not only in math and science, but also in reading and writing, and while I appreciate being kept on my toes, it can feel a little icky to attempt a problem in good faith only to find that the premise is flawed. In the classroom, I try to frame that as a moment when we get to be "smarter" than the book--and if I know there's an issue, I fix it for the students ahead of time or specifically direct them to find the mistake. Doing otherwise produces a "gotcha" situation, which is not a productive learning environment.

Thanks! You pegged my thoughts, though I hadn't thought of the "Car Talk" comparison.     

You're right, those "gotcha" moments aren't productive.    I work hard to get students not to be intimidated by a problem that seems impossible and wrong.   "Even I look at a problem on first read and say, "what ?!?"" ... but that it's what we do next that matters... figure out what we *do * know about the problem, draw a picture, etc.   When students realize that for problem X that wouldn't have done any good because it was a flawed problem... it makes it really hard to have faith that future problems will work out.   (Fortunately, I guess, on an actual test like the GED you never know that it was a flawed problem... ) 


I think that many problems that appear on published tests and review sheets are written badly and are often ill-conceived.  It may be that they are written by persons who don't have a clear idea of the mathematics involved and/or are muddled about the purpose of the problem or don't have a reasonable sense of what the student is supposed to know at that point in their education.  Also, because of the recent push to include the idea of drawing out results from data, we see many problems with little lists/tables of values of some function, often with unclear context/labeling and leading to confusing and sometimes  nonsensical problem statements.   

In the case of Challenge Problem - May 22, it would set the context more clearly if the problem was introduced by saying: "We will consider two parallel lines 4 inches apart and all parallelograms and triangles with a bottom base side on one of the lines and the top of the figures on the other line".  Evidently they thought maybe students didn't know how to find the area of a parallelogram, so they would remind them by giving the area formula A = 4*L where L is the length of the base in inches, and they are expecting the student to use the formula to calculate some areas..  It looks like they want to find out if the student does know, or can figure out from the data, a formula for the area M of a triangle with the same base length.  So they give a table of values of M for various base lengths p. My guess is that they hoped the student would make a third column giving values of A for these same base lengths p.  (Perhaps they thought using p rather than L would make it a bit less trivial.)  And then it would be obvious that M is 1/2 of A. 


For the purpose of teaching students about how to compute areas, I think the context above is a very good one.  Start with a straight "paper strip",  "ribbon", "sidewalk" or "road" of constant width W, that is, a section of the region lying between two parallel lines W units apart, and touching each line.  Assume each end of our strip results from a straight line cut across the strip.  There is a natural center-line of our section (half way between the parallel bounding lines) and we will let L denote its length.

Theorem.  The area of our strip is W*L no matter what angles the end cuts make with the base. 

This takes a bit of proving.  In the easy cases we can just make a new cut perpendicular to the base and through the center point of one end of the strip and then swivel the snipped off right triangle 180 degrees around the center point to get a new strip with that end perpendicular to the base ( now a square end) and still having the same center line length L as before.  And if the other end is a good case as well, we can do the same to that.  So our original strip has the same area as the new rectangular strip we have produced, and this rectangle has area W*L.  You can easily see what to do in the harder cases by adding a good strip onto an end and later subtracting areas..

Notice that this theorem gives one formula that computes the area for all trapezoids and triangles.  But now generalize to constant width borders around buildings and sidewalks, as long as they are made up of straight segments (trapezoids meeting at corners at whatever angle).  Just run your measuring tape along the centerline to measure L and then multiply by W to get the area.

Now generalize further by chopping up road curves and sections of disks or rings (annular region between concentric circles), into very thin slices that are very nearly triangles or trapezoids (our cuts should be perpendicular to both sides of the road).  We can still measure the centerline length L, and passing to the limit the area is still W*L. So to find the area of a section of road about to be paved, just measure the centerline and multiply by the constant width.

So we have a very easy formula to use to calculate the areas of many common types of regions.  It's a lot easier and more precise than trying to directly use the definition of area of a planar region R as the maximal number of 1X1 stamps that you can paste onto R without overlapping and without going outside of R, or equivalently, the minimal number of 1X1 stamps that you can paste on R to completely cover R.  (Of course in both cases you may cut up stamps into tiny pieces if necessary.) 

I think you put into words what the 'desmos" drawing described visually.   -- as well as that this problem is ill-conceived and poorly written ;)   

It's great when we can get students to understand the concrete, "stick these stamps on it" part .... and it can be approached at different levels.   Some students pretty much 'get" that a triangle can be cut up  and shown to be a set of half-rectangles... and others take the knowledge further into other shapes.   My minimum requirement is that they get past "area is length times widthe, right?"   (especially when a problem has asked for perimeter... and it happens...)


One of the nice outcomes of the discussion above about constant width paths where the area is always  W*L, is that it works so neatly for a ring region S between concentric circles with radii b<c.  The usual formula for computing  that area is  Area(S) = pi*c^2 -- pi*b^2 , which you get assuming that you already know the formula for the area of a circle .   But notice what we get if we factor this as

Area(S) = pi* (c^2 -- b^2 ) = pi* (b+c) * (c-b) = (2*pi)* ( [b+c]/2 ) * (c -- b )

There we see the constant width of the ring region is ( c-- b ), and since the centerline of this region is a circle whose radius is [b + c]/2  then the length of this centerline circle is  2*pi* ([b + c ]/2).   Or you might note that the diameter of the centerline circle is [b + c ], so its length is pi* [b+c].  And of course when b = 0 we get the usual formula for the area of a circle.  These formulas are special cases of the generalized theorem discussed above. 

For a concrete version of this you might think of a curvy path with no sharp corners and a person walking along the centerline of length L dragging a broom of width W that is full of paint.  The painted area  then has area L*W.