Below are pictures of an equilateral triangle, a square, a regular hexagon, and a circle each having the same perimeter:
In order to use the formula, Area = \(\frac{1}{2} \times\) Base \(\times\) Height for a triangle, we need to choose a base and then determine the height of the triangle with this base. Labelling the vertices of the triangle \(A\), \(B\), and \(C\), let \(M\) be the point where the perpendicular line from \(A\) meets \(\overline{BC}\):
Note that \(|AB| = |AC| = |BC| = \frac{1}{3}\) since the perimeter of triangle \(ABC\) is \(1\) and all three sides are congruent. Since \(\overleftrightarrow{AM}\) is perpendicular to \(\overline{BC}\) we can apply the Pythagorean theorem to the right triangles \(AMC\) and \(AMB\):
We know that \(|AC| = |BC| = \frac{1}{3}\) so these two equations show that \(|CM| = |BM|\). Since \(|BC| = \frac{1}{3}\) this means that both \(|AM|\) and \(|BM|\) are \(\frac{1}{6}\). We now apply the Pythagorean theorem again to triangle \(AMC\) to find \(|AM|\):
So \(|AM|^2 = \frac{3}{36}\) and $$ |AM| = \frac{\sqrt{3}}{6}. $$ Now we have found the height of triangle \(ABC\), using \(\overline{BC}\) as base, and so the area is $$ \frac{1}{2}\times \frac{1}{3} \times \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{36}. $$
A regular hexagon with perimeter \(1\) can be divided into \(6\) equilateral triangles, each having side length \(\frac{1}{6}\) as pictured below.
We have found in part (a) the area of an equilateral triangle whose side length was \(\frac{1}{3}\) unit. We can repeat the argument from part (a) to find the area of each of these triangles and will find that it will be \(\frac{1}{4}\) of the answer from part (a) as the base and height of the equilateral triangle from part (a) have both been scaled by a factor of \(\frac{1}{2}\). So the area of the hexagon will be \(\frac{6}{4}\) times the area of the equilateral triangle in part (a): $$ \text{Area(Hexagon)} = \frac{6}{4} \times \frac{\sqrt{3}}{36} = \frac{\sqrt{3}}{24}. $$
Commentary
This problem is part of a very rich tradition of problems looking to maximize the area enclosed by a shape with fixed perimeter. Only three shapes are considered here because the problem is difficult for more irregular shapes. For example, of all triangles, the one with fixed perimeter \(P\) and largest area is the equilateral triangle whose side lengths are all \(\frac{P}{3}\) but this is difficult to show because it is not easy to find the area of triangle in terms of the three side lengths (though Heron's formula accomplishes this). Nor is it simple to compare the area of two triangles with equal perimeter without knowing their individual areas. For quadrilaterals, a similar problem arises: showing that of all rectangles with perimeter \(P\) the one with the largest area is the square whose side lengths are \(\frac{P}{4}\) is a good problem which students should think about. But comparing a square to an irregularly shaped quadrilateral of equal perimeter will be difficult.
For this problem, very explicit shapes have been chosen aiming at providing an opportunity for students to practice using their knowledge of different geometric formulas for area and perimeter. The teacher, in order to reinforce part (d), may wish to draw examples of a hexagon and octagon (both sharing the same perimeter as the other figures). The central idea is that as the number of sides of the polygon increase, the polygon looks more and more like the circle and in particular its area is getting closer to the area of the circle.
Note that the last part of part (e), asking students to think about the regular pentagon, is designed to help them observe that as the number of sides of the regular polygon increases, the polygon looks more and more like a circle. The results of (a), (b), (c), and (d) also might lead to the conjecture that as the number of sides of the polygon (with fixed perimeter) grows, the area increases. While students are not in a position to verify this conjecture, it is good for them to think about the problem. The teacher may also wish to share an interesting historical context of this problem which is described here:
Dido's Problem
Finally, effort is made in the solution to develop numerical sense in comparing the size of the areas found in parts (a), (b), (c), and (d). If the teacher is interested in stressing this aspect of the task, then it would be appropriate to instruct the students not to use a calculator when doing part (e) of the task. If this aspect is not important, then the answers to (a), (b), (c), and (d) can simply be compared with the aid of a calculator.