The concept of area – Part 3

This is the third part of a multi-part blog post on the concept of area.

Pop quiz!

What’s the formula for the area of a triangle?

If you answered ``area is one-half base times height’’ or \(A = \frac{1}{2}bh\) (newer browser needed to see math rendering) or something of the sort, give yourself an ``A’’.  It wasn’t a trick question.  (Not challenging enough?  More math problems below.)  That’s the formula that everyone learns and hopefully knows, but if you’re in high school and can’t remember it, don’t worry if you are taking the SAT; the College Board provides the formula along with several others on a ``cheat sheet’’:

We here at ccssimath.blogspot.com were curious to whom this triangle area formula is given on state level exams.  We called several education offices and some were not so forthright, citing confidentiality (for a reference sheet?!?), but others have online images of the pages that are provided.  Our list below is representative, but not exhaustive (and we took some liberties with chopping up the images for clarity).

New Jersey provides the formula on GEPA, its 8th grade state-wide assessment:

as well as on HSPA, which is taken by 11th graders:

Florida gives it to 6th through 8th graders on FCAT:

So does Indiana on ISTEP+:

The New England NECAP offers this variation in the 6th Grade:

Massachusetts’ MCAS uses similar formula sheets for Grades Six through Ten, but in the 5th grade, they provide the more explanatory:
We were somewhat surprised by the variability:
  • grade levels at which formula is given (some states never provide it)
  • if formula is given, it is accompanied by an acute triangle, right triangle, obtuse triangle, or no picture
  • height drawn or not drawn as a dashed line
  • multiplication indicated by ⤫, • or juxtaposition
  • amount of explanation given along with formula

So what are we driving at?  That a common standard is good?  Well, sure, inconsistencies in mathematics notation are never helpful, but students arguably shouldn’t need these ``cheat sheets’’ anyway; they should know how to find the area of a triangle.

The same goes for the Pythagorean Theorem, area of a circle, etc.
But we digress.

One element every formula above has in common is the basic arrangement:
\( \frac{1}{2}\) multiplied by b multiplied by h.

It is this common format that we here at ccssimath.blogspot.com take exception to.  Give us leave to explain.


How and when does CCSSI address the topic of the area of a triangle?  The topic is first mentioned in 6.G.1 which states, ``Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes...’’

Before Grade 6, the only ``advancements’’ CCSSI has made in the topic of area since Grade 3 is 4.MD.3’s ``...find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor’’ (a problem that belongs back in the 3rd grade along with area of rectangles, without attempting to use algebra terminology prematurely and unnecessarily), and 5.NF.4b which extends the area of a rectangle to include fractional lengths.  We are not impressed.  We don’t consider rehashing the rectangle area formula A=lw to use fractional lengths a useful application of fractions, and neither 4.MD.3 nor 5.NF.4b extends the concept of area or formally introduces anything new.

The concept of area that we previously characterized as critically important should be steadily advancing; not in the holding pattern between Grades 3 and 6 that CCSSI is keeping us.

In its 2000 Mathematics Curriculum Framework, Massachusetts had included the area of triangles (by counting) in Grades 3-4, and by formula in Grades 5-6, a possible improvement that CCSSI could have considered.

However, the triangle area formula requires multiplying a fraction by a whole number, an ability that CCSSI puts either in 4.NF.4 ``Apply and extend previous understandings of multiplication to multiply a fraction by a whole number.’’ or in 5.NF.4 ``Apply and extend previous understandings of multiplication to multiply a...whole number by a fraction.’’

At first glance, we couldn’t determine how CCSSI is distinguishing the product of a whole number and a fraction in these two distinct standards which are separated by a year.  Then we realized (through its illustration) CCSSI intends in 4.NF.4 ``multiply a fraction by a whole number’’ to mean the whole number should precede the fraction, e.g., \( \frac{5}{4} = 5 \times (\frac{1}{4}) \), and in 5.NF.4 ``multiply a...whole number by a fraction’’ to mean the whole number should follow the fraction.

We are not going stop here to discuss in detail our opinion of CCSSI’s treatment of fractions (we’ll do that in another blog post), but under CCSSI’s scheme, it follows that the formula \(A = \frac{1}{2}bh\) must come sequentially after 5.NF.4.  (However, the formula is never explicitly mentioned, something we will return to shortly.)

This is why CCSSI must either put the area of a triangle later in Grade 5 (which it doesn’t) or in Grade 6 (which it does).

In 6th grade, when students are being introduced to negative numbers and their operations and basic algebra operations, the area of a triangle seems quite pedestrian in comparison.  In fact, it is even orphaned from fractions.  We think the area of a triangle is not so complicated and ought to come earlier, so that the topic of area does not lie fallow for two years in students’ brains.  However, there’s going to be a clash of misaligned standards...IF we don’t intervene.

Therefore, we here at ccssimath.blogspot have a modest proposal (with no offense to Jonathan Swift).

Have you ever gone past the same place or looked at something many times before noticing some particular detail?  This experience is common, of course (and www.urbandictionary.com has tried to coin a term ``vu jàdé‘’ (definition 3)), but we are suggesting something that perhaps the entire mathematics Establishment from mathematicians to state education departments to curriculum ``experts’’ to NCTM to teachers have missed.

The formula for the area of a triangle should be learned as

\[ A = b \times h \div 2 \]
Now, of course, we know purists may object to this mathematical sleight-of-hand. (``It’s the same thing!’’)

In fact, it’s better for several reasons.
  1. Why use fractions when you can divide instead?
  2. By presenting the formula this way, without fractions, the concept of the area of a triangle and its formula can be learned before fractions.  Yes, we know that if the base and height are both odd numbers, then the answer will be a fraction, but it adds nothing to the concept to consider every possible combination of base and height, whether they are both odd, or even fractional or any other kind of number.  Just as we said before it’s silly to review the area of a rectangle using fractional lengths, because it adds nothing to the concept but simply allows posing of a mixed number multiplication exercise in a picture of a rectangle, so too, can we simply make sure that either the base or height be an even number to avoid initial complications with the triangle area formula.  The area of a rectangle can reach closure with only whole numbers, and so too can the area of a triangle.
  3. Learning the formula this way introduces the concept of a fraction as a multiplicand (or multiplier) before fraction multiplication is formally introduced.  We have said before that understanding a concept should precede formal introduction.
  4. Changing the triangle area formula also makes it an application of division.  In third grade, students learn division by single digit numbers (see 3.OA.3, 4.NBT.6, et al.), and what application of this skill is more basic and obvious than using it to calculate the area of a triangle by finding a product and in the last step, dividing by two?  Getting in the habit of dividing numbers by two is a valuable life skill, in and of itself, both on paper and mentally, and triangle problems create an obvious situation in which to practice.
  5. Putting the area of a triangle in the third grade (which we advocate), right after the area of a rectangle, is repeating a practice that we’ve suggested before: multiple applications of a concept abstracts the concept.  We have already lambasted CCSSI for teaching that area is a bunch of squares.  Our proposal helps define a broader understanding of area: now students will be able to decompose not only rectilinear shapes, but any polygonal shape into rectangles and triangles, and solve the problems with multiplication, division, addition and subtraction.  And it will enable students to apply estimation skills more accurately in ``approximate area of a lake'' problems.
  6. Lastly, knowing how to find the area of a triangle in third grade will open an entire realm of interesting problems to solve, which contain our triumvirate of attributes: length, connectivity and dimensionality (examples to follow).


Consider first the area problems students typically see, and the tradition that CCSSI is continuing in 6.G.1.  What kinds of polygonal shapes are presented as area problems most often?

We’re trying to keep these blog posts ``concise’’ (HA!), so we’re avoiding in this post a more general discussion of the area of quadrilaterals, but to give the reader a preview, the following problem was given on the 2005 NAEP:

To answer this question (we won’t deign to say ``solve’’), students needed to be able to measure with the centimeter ruler that’s provided (second grade skill, see 2.MD.1), students needed to plug those numbers into the formula for the area of a parallelogram, A=bh (provided on the NAEP reference sheet, and useful, so long as you knew what base and height meant), multiply two numbers (third grade), and round the answer to the nearest 10 (second or third grade, see 3.NBT.1).  Even with all of these necessary skills, this question lacks any depth whatsoever; there’s no mystery what must be done to get to the end.

But we digress from triangles.  (Oh, this picture could be decomposed and/or recomposed using triangles, you say?  Well, sure, that’s why this question is in this blog post, but how many students made either connection?  It looks like a parallelogram, and judging from the following results, was treated like it came from Mars.)

38% (not a typo) of graduating seniors (not a typo, either) got the correct answer, C, so Houston, we have a problem.  (Yes, we read the Brookings opinion of the disconnect between NAEP and common core grade level standards.  That’s a separate debate.)


Consider, as an alternative, the following question:
On its face, it already appears more challenging than the NAEP problem because the shaded quadrilateral is not a rectangle, square, parallelogram or trapezoid for which the area formulas were provided.  That means it’s not going to be answered by plugging numbers into a formula.

This problem has the length, connectivity and dimensionality that we are espousing, but not for 12th graders; we think it is appropriate for fourth or fifth graders.

  • First, this problem is longer than the 12th grade NAEP problem.
  • Second, this problem has connectivity; you need to know things and apply thinking skills.  To solve this problem requires some insight: it’s necessary to draw an ``auxiliary’’ line, or at least to imagine it.  This problem requires knowing the formula for the area of a triangle (yes, that’s why it’s also in this blog post), so it requires multiplication and division.  And it requires adding the areas of two separate regions that are conjoined rather oddly.
  • Finally, this problem has dimensionality: the way to proceed and the end are not so clear at first glance.  The ``auxiliary line’’ step is not immediately apparent; the problem provides no hints.  
There’s nothing in this problem that is beyond the ken of elementary school students; it doesn’t even require fractions (now that we’ve revised the area of triangle formula).  It requires understanding whole numbers, the operations of addition, multiplication and division, plus some actual thought.

To summarize, we at ccssimath.blogspot.com continue to advance the cause of a well thought-out sequence that understands where things belong, understands where you are coming from and where you are going, and poses the right problems to foster the real thinking processes that we so strongly believe are the hallmarks of an effective education.  Reformulating and shifting the area of a triangle from 6th to 3rd grade is one example of a major rethinking that opens a world of problem solving that is both challenging and age appropriate.  Although it is just one small step in useful curriculum reform, it promises to help raise American education well past what CCSSI envisions.

See this additional problem.


  1. Mathius197811 April, 2013

    Thank you for posting your series. For those of you stuck on the problem, look in the math subreddit for a discussion of this. I am not posting a link to encourage you to spend some time trying to solve it on your own.

    At the author, glad you found my reddit post. I am sorry if you did not want an answer posted. Feel free to delete this if you do not want a guide to out there.

    Again thanks

  2. Mathius197811 April, 2013

    On 5 triangles how do you solve problem 59 in a way that 6-8 yo can understand and account for the length of the curve point 0 traces as it goes around a corner?

    1. This is the problem the commenter refers to:


    2. Cleverness is required, but the point is that in going around all three corners, 360 degrees will be traversed. So the total distance there is the circumference 2*pi*r . During that pivoting, the role of the circle and its center switch, where the center rotates around the point on the circle that is "stuck" to the corner of the triangle. This might be hard to explain to a 6-8 y.o. but it may not get much easier as that student gets older, if he or she has trouble visualizing in two dimensions.

      So the total distance traveled by the circle's center is the perimeter of the triangle plus the circumference of the circle.

  3. The five triangles problem is nice, a perfect sort of problem for seeing if the student understands how and why the (1/2) b h formula works.

    The one given above on this page is a bit harder, you have to draw in the diagonal from upper left to lower right, then you have (1/2) b h for each of the two resulting shaded triangles so the answer is 3x8/2 + 5x6/2 = 27 cm^2. This is more like a Math Olympiad problem than one I would expect to see in class, certainly not for a non-gifted class. It's nice to talk about encouraging thinking, but if you can actually get average students to be able to do such problems, you are a remarkable teacher!