Area = Perimeter

Why, oh why, are these topics taught so often at the same time? All that does is build in a confusion As a secondary specialist I would point to the distinction in units—perimeter, linear length, metres or similar for units; area, square units, probably square metres, possibly hectares.

It is true that in practical terms the combination of these two measures occurs at much the same time. Even you are sure you and your class have these terms very much not confused, please make sure you use a defining unit to be clear at all times which is which.

Adding to the confusion for those determined to confuse, and so used by difficult people like me to ascertain if there is confusion at all is a problem such as this one:


1.  Can you find a rectangle whose perimeter is numerically equal to its area?

Well yes you can, and quite quickly. If you assume that only integer solutions are wanted—maybe you did already— this also doesn’t take long.

Once you have found one such solution, can you find another rectangle fitting the rule? Are there any more?


  1. 2.What is the maximum area you can enclose with a fixed perimeter?

This can be made more approachable like this:

2.0 I can buy wire open mesh fencing as I use for keeping sheep in 50m lengths. What is the most area I can enclose with one roll of fencing?

2.2 I decide to make use of an existing straight boundary wall to build a rectangular enclosure. I start by planning a square enclosure but then look at rectangular solutions. Which gives me the biggest area?

2.3 Find the best triangular solution. If this is too general for you, try exploring some specific triangles and use this experience to guide yourself to an optimum solution.

2.4 By trying more polygonal approaches, find the absolutely best solution, commenting on this.

2.5 It would be more sensible (in the sense that I enclose more area with the addition of the same roll of fencing, if I use more of the existing boundary fencing. How do my answers change if I use an existing right-angled corner with straight sides? a) a straight run, b) making a rectangle, adding two sides c) adding three straight sides d) going for broke (you should interpret that, based on the previous answers)


The same questions work well with middle school, where drawing solutions and then perhaps using approximate methods is a good approach. This encourages investigation and teamwork. Showing that four straight lengths encloses more area than either of the earlier solutions. The more able can be given the most practical work, by you deciding how often posts need inserting, whether all posts are the same (they are not, if wooden; you need a bigger and more expensive post and strainer for a change of direction) and so you can work the able towards a proper farmer’s solution, the most area for the least expense either (i) money spent, labour costs irrelevant or (ii) pricing the labour too. This could (can, indeed) extend to a whole week’s project. If the questions are posed by the class from discussion as a whole, then the class as a whole or in groups can find answers to its own questions. By making the project their own, they will discover far more than if driven by teacher. If the teacher encourages different approaches (e.g. by grouping students together) the class as a whole can benefit. Opportunities here for presentation of discoveries, for ICT solutions, for interaction outside school (“So how do you do this sort of fencing? Can you find out?”; “Is there an extra cost if you add another corner in the fence?” “Do you need a gate? Well wouldn’t that be a fixed extra cost and extra length that effectively makes the roll of fencing go further?” “Is the fencing relatively cheap so some of this is a silly problem? Okay, then let’s look at the problem if we instead are looking at dry stone walling; look at the problem in general and then tell me about choices of fencing material. Maybe you’re looking for the cheapest way to enclose an area, in which case what we fence with and how we construct the fence is the issue. What is the purpose of the fence, really? Does our answer change if we switch from sheep to cattle? Horses? Pigs? Llamas? Deer? Goats? Wheat? Does our answer change if we consider how long we want the fence to last? Did we switch from maths to geography somewhere here? And to the budding economist: “No, right now your time is very cheap and you’re solving the problem in general, so everyone you tell benefits” Or sell the results to, indeed.


Some classes can do this only in class. Some would find it stretching as homework; which approach fits your class depends how well you have trained them to think for themselves. Consider rewriting the questions I pose so as to lead them through the exploration. I suggest using such an approach as general training in the use of maths, but I’d also suggest setting a time limit (“See how far you get in an hour”), for this is the sort of work that some will put large amounts of time into; where that is their choice without deleterious effects on other work, wonderful, otherwise it comes back to bite and a few will lose heart. Which brings you back to doing such things in class, where you can walk around encouraging exploration.


DJS 20160621


Partial answers / results:

1. Let the rectangle be of dimension a, b. Then obviously P= 2(a+b) = ab = A. Also, if (a,b) is a solution, so is (b,a) so I choose to disregard those immediately, choosing a<b.

Now let a =kb, so k>1.  Somehow one thinks k is integer.

Why is that? [Because diagrams were drawn on squared paper, in the case of the classes I tried this on].


So now 2a(1+k) = a2k so a≠0 gives 2(1+k)=ak and, after rearranging,  k = 2 / (a-2).


If a and k are integers, then there are only two solutions (Y12/3 challenge: why is that?):

k=1 and k=2, which means the rectangle has dimensions 4x4 and 3x6 (and 6x3). Sizes in linear units, I don’t care what. And the zero solution, (0,0), of course.

Can there be non-integer solutions?

Choose a value for a [we have already found the solutions for a={3,4,6}]  Solution pairs (a,b) all fall into the general form where b = 2a / ( a-2),  so a>2 and so we have a set of solutions such as (3,10/3), (5,14/5)  (integer values of a) and :-

I conclude that there is an infinite set of solutions for any a>2, b>a, with aϵ𝓡 (a real number).


  1. 2.All you’re doing here is fixing the perimeter P = 2(a+b) and maximising the area A=ab.

2.1 A Y12 solution here is to write A=a(P/2-a) and then dA/da = P/2 -2a, suggesting that the rectangular solution is when a=P/4, b=P/2, A=P2/8. Of course most Y12 will demand there is an x in the problem. Smart students will explore shapes beyond the rectangle and may well discover that symmetry is a benefit. With no wall to build against, the maximum enclosed area comes from a circular enclosure (equally weak at all posts, all tending to be pulled inwards and all really needing strain wires outside or extra slanting posts inside). It should be helpful to show this value as an upper limit to increasingly polygonal solutions.


2.2 P=2a+b, A=a(P-2a), dA/da = P-4a => maximum at a=P/4, A= A=P2/8


2.3 Finding the triangular solution is interesting at Y12 too. The traditional triangle has sides {a,b,c}, c is along the boundary and the included angle C between a and b, so the area is 1/2 ab sinC. This will have a maximum at C=π/2 (Y12, use radians not degrees) and a few will realise or rediscover that the altitude of this triangle, once found, doesn’t change the area (which means a=b is a solution).

Some will show (or decide) that tan A = a/b and a few will see that a=b and A=B=45º=π/4 is a maximum, but will not try to prove that. Differentiation shows that a solution for a=b occurs when a=P/2, A=P2/8 but really the maximum distance from the boundary is P/2√2 and there is a large solution set. The problem, while apparently very simple, has a host of potential pitfalls for the budding mathematician (engineer, scientist, rational thinker, future consultant of some sort) to explore—and several p[laces where the learned theory comes into question, such as what one might to with too many variables in a problem to be differentiated.

Some will explore, after encouragement, what happens if there is a right angle in the solution (“We can’t do that, but we can do this less difficult problem, so let’s try it”). That gets a=b=P/2, A= A=P2/8 and that in turn may lead to some better understanding. Again, perhaps with encouragement.

A few will connect (possibly as a revelation) that last solution set, the envelope of solutions with C=90º, with the curved solution, while some will have difficulty with A=P2/8 being the solution both times so far. If someone decides that a regular solution is optimum (or merely less difficult), they might progress to a circular arc as solution.


2.4 P=πr, A=πr2/2 = π/2   (P/π)2 = P2/2π  which is satisfyingly bigger than A=P2/8 but not possible to build as fencing (though it is build-able as wall).


2.5 if we use two straight existing walls with a right angle corner, then a single straight line gives, I think, maximum area when symmetrical so the two equal wall sides are of length a=  (1-√5) P/2. Please check that, since I was rushing.

Adding two sides to make a square, lengths are all P/2 and area P2/4.                       clearly 0.25  P2

Adding three equal sides of an octagon makes the area (I think) A= (5/4+√2) P2/9 , approx  0.296 P2

Adding a quadrant of fencing ( circular arc) has radius 2P/π and area P2/π               approx 0.318 P2



A natural extension would be to wonder about adding a dimension:

  1. 3.Is there an integer value cuboid whose area is numerically the same as its volume? What about other regular solids? 

I think (rapid scribbling, no checking) a tetrahedron has height 12, [4A=Ah/3], the cube has side 6 and the sphere has radius 3. To have a solution set really we should be listing the radius equivalent and a decision needs to be made whether that is the long or the shorter values, or both. Trying to do that, then {name, No. faces, side, short radius, long radius} has (sphere, infinite, 0,3,3), (cube, 6, 6, 3,√3/2) and (tetrahedron, 4, 4√6, 4√(2/3), 8√(2/3)). I’m not certain about that last set, it needs a separate study of the tetrahedron first.


Suggested Exercise: A regular tetrahedron ABCD has side a. let L,M,N be the midpoints of BC, CA and AB respectively. Let H be the point on face ABC where AL meets BM, opposite vertex D. Pick points E,F,G to be opposite A,B,C respectively. Let O be the central point where AE meets DH.

  1. (i)Sketch the tetrahedron and label the points.                                                                [2]
  2. (ii) By considering triangle ADL (the midpoint of AD is P) or otherwise, find lengths AL and DL, and show that AH=2HL    [2,5] (if pushed for time, do the proof by declaring where medians meet for 2/5)
  3. (iii) By considering triangle ADL, find length DH                                                              [3]
  4. (iv) Attempt to show that DO=2OH and state the value of each.                                     [5,1]   total 18.

OH is the least distance from O to a face, DH is the distance to a vertex.


If using vector methods instead of trigonometry, this might be another useful exercise

Let ABCD define a regular tetrahedron: the directed lines OA, OB, OC, OD are unit vectors a,b,c,d respectively of equal length. Using the labelling of the previous question,

  1. (i)show that AL is b/2+c/2-a, declare DL in a similar way and
  2. (ii) find |DL|


© David Scoins 2017