Geometry · 7.G.B.6 · G.MG.A.1

Area of a regular polygon

A regular hexagon has no length and no width to multiply. How do you measure the space inside it?

You need

The apothem of a regular polygonGeometry Area of trianglesMath 6

Where you are 15 in Area and perimeter

▴ 10 above Area of triangles Area of any polygon by cutting it into triangles and rectangles Area of composite figures and shaded regions Why the circle area formula works

Area of a regular polygon

When you reach for it

A rectangle’s area is length times width, and a triangle’s is half the base times the height. A regular hexagon has neither a length nor a width, and neither does a regular pentagon or octagon. Yet these shapes are everywhere: floor tiles, the head of a bolt, a road sign, the cells of a honeycomb, the floor of a gazebo.

Reach for this move whenever the shape is a regular polygon, one whose sides are all equal and whose angles are all equal. The move is to cut the polygon from its center into equal triangles, find the area of one of them, and multiply by how many there are. The height of every one of those triangles is the polygon’s The apothem of a regular polygon, which is the reason the apothem is worth naming at all. Done once with letters, the move gives a single formula for every regular polygon:

A = \frac{1}{2}aP

half the apothem times the perimeter.

The procedure

Cutting from the center

Every regular polygon has a center: a point equally far from all of its vertices and equally far from all of its sides (see regular polygons and their circles). Draw a segment from the center to every vertex. A polygon with n sides is cut into n triangles, one standing on each side, and they fill the polygon without overlapping.

Figure 1Five congruent triangles, each standing on a side. Each has base s and height a.

The triangles are all congruent. Each has two sides running from the center to a vertex, and these all have the same length r, the radius, because the center is equally far from every vertex. Its third side is a side of the polygon, of length s, and those are all equal because the polygon is regular. Three sides of one triangle match three sides of another, so the triangles are congruent by side-side-side. Congruent triangles have equal areas, so finding one is enough.

Each triangle’s height is the apothem. Take the polygon’s side as the triangle’s base. The triangle’s height on that base is the perpendicular distance from the opposite vertex, which is the center, to the line of the base. That perpendicular distance from the center to a side is exactly what the apothem a is. So each triangle has area \displaystyle \frac{1}{2}sa.

Add the n triangles.

\begin{aligned} A &= n \cdot \frac{1}{2}sa \\ &= \frac{1}{2}a(ns) = \frac{1}{2}aP \end{aligned}

The middle step only reorders and regroups the factors of a product, which the commutative and associative properties of multiplication allow. Then ns is n sides of length s laid end to end, which is the perimeter P. Working triangle by triangle keeps the picture in view, and \displaystyle \frac{1}{2}aP is quicker once you trust it. They always give the same number.

A check on a shape you already know. A square of side 10 has its center 5 from each side, so a = 5 and P = 40:

A = \frac{1}{2}(5)(40) = 100

which is 10^2, as it has to be.

When the apothem is not given

Often you know the side, or the radius, and not the apothem. The apothem meets its side at a right angle, and because each slice is isosceles (two of its sides are radii), it meets the side at its midpoint. So the apothem cuts every slice into two right triangles, each with legs a and \displaystyle \frac{s}{2}, hypotenuse r, and an angle of \displaystyle \frac{180^\circ}{n} at the center, half of the central angle \displaystyle \frac{360^\circ}{n}.

Any two of these, at least one of them a length, fix the rest. With two lengths known, the Pythagorean theorem gives the third:

a^2 + \left(\frac{s}{2}\right)^2 = r^2

For three shapes the angle does most of the work.

The square. The angle at the center is 45^\circ, so the half slice is an isosceles right triangle and \displaystyle a = \frac{s}{2}.

The regular hexagon. The central angle is 60^\circ. Each slice is isosceles, so its other two angles share the remaining 120^\circ equally and are 60^\circ each. The slices are equilateral, so r = s, and the apothem is the height of an equilateral triangle of side s:

a = \sqrt{s^2 - \left(\frac{s}{2}\right)^2} = \frac{\sqrt{3}}{2}s

The equilateral triangle. The angle at the center in the half slice is 60^\circ, so the angle at the vertex is 30^\circ. A right triangle with a 30^\circ angle is half of an equilateral triangle (put two copies back to back along the longer leg and all three angles are 60^\circ), so the side opposite the 30^\circ angle is half the hypotenuse. That side is the apothem: \displaystyle a = \frac{r}{2}.

For any other n, one length and the angle at the center give the rest through right triangle trigonometry. The leg \displaystyle \frac{s}{2} is opposite the angle \displaystyle \frac{180^\circ}{n} and the leg a is next to it, so

\begin{aligned} \tan\frac{180^\circ}{n} &= \frac{s/2}{a} \\ a &= \frac{s}{2\tan\left(\frac{180^\circ}{n}\right)} \end{aligned}

Worked examples

Example 1

Side and apothem given

A regular pentagon has sides of 4 cm, and its apothem measures 2.8 cm to the nearest millimeter. Find its area.

Solution

The perimeter is P = 5 \cdot 4 = 20 cm. With the apothem known, the formula applies at once:

A = \frac{1}{2}(2.8)(20) = 28

so the area is about 28\ \text{cm}^2.

Triangle by triangle gives the same: there are five slices, each with base 4 and height 2.8, so each has area \displaystyle \frac{1}{2}(4)(2.8) = 5.6, and 5 \cdot 5.6 = 28.

The apothem was rounded, so the area is too. The true apothem lies between 2.75 and 2.85, so the true area lies between \displaystyle \frac{1}{2}(2.75)(20) = 27.5 and \displaystyle \frac{1}{2}(2.85)(20) = 28.5.

Example 2

Finding the apothem of a hexagon

Find the exact area of a regular hexagon with sides of 6 cm, and give it to two decimal places.

Solution

The apothem is not given, so find it first. The central angle is \displaystyle \frac{360^\circ}{6} = 60^\circ, and as in the procedure the six slices are equilateral triangles with all sides 6.

Figure 2A regular hexagon of side 6 cuts into six equilateral triangles. The apothem is the height of one of them.

The apothem is the height of one of them. It meets the bottom side at its midpoint, splitting it into 3 and 3, and the Pythagorean theorem in the right triangle with hypotenuse 6 gives

\begin{aligned} a^2 + 3^2 &= 6^2 \\ a^2 &= 27, \qquad a = \sqrt{27} = 3\sqrt{3} \end{aligned}

The perimeter is 6 \cdot 6 = 36, so

A = \frac{1}{2}\left(3\sqrt{3}\right)(36) = 54\sqrt{3} \approx 93.53

The area is \displaystyle 54\sqrt{3}\ \text{cm}^2, about 93.53\ \text{cm}^2. Keeping \displaystyle \sqrt{3} until the last line means the only rounding is the final one.

Example 3

Working backwards to the apothem

A regular pentagon has a perimeter of 30 cm and an area of 61.9\ \text{cm}^2. How far is its center from each side, to the nearest hundredth of a centimeter?

Solution

The distance from the center to a side is the apothem. The formula \displaystyle A = \frac{1}{2}aP ties the three quantities together, and two of them are known, so solve it for the third. Multiplying both sides by 2 and dividing both by P gives \displaystyle a = \frac{2A}{P}:

a = \frac{2(61.9)}{30} = \frac{123.8}{30} \approx 4.13

The center is about 4.13 cm from each side. No angle and no side length were needed: area and perimeter alone fix the apothem.

Example 4

An equilateral triangle in a circle

An equilateral triangle has its three corners on a circle of radius 12 cm. Find its area.

Solution

The circle’s center is the triangle’s center, since it is equally far from all three corners, so r = 12. The central angle is \displaystyle \frac{360^\circ}{3} = 120^\circ, and the half slice has a right angle at the midpoint of a side, 60^\circ at the center and 30^\circ at the corner. As in the procedure, it is half an equilateral triangle, so the apothem, opposite the 30^\circ angle, is half the hypotenuse:

a = \frac{12}{2} = 6

The other leg is half a side. By the Pythagorean theorem,

\begin{aligned} \left(\frac{s}{2}\right)^2 &= 12^2 - 6^2 = 108 \\ \frac{s}{2} &= 6\sqrt{3}, \qquad s = 12\sqrt{3} \end{aligned}

So \displaystyle P = 3 \cdot 12\sqrt{3} = 36\sqrt{3}, and

A = \frac{1}{2}(6)\left(36\sqrt{3}\right) = 108\sqrt{3} \approx 187.06

The area is \displaystyle 108\sqrt{3}\ \text{cm}^2, about 187.06\ \text{cm}^2.

Check with the triangle’s own base and height. The line from a corner through the center to the midpoint of the opposite side is the triangle’s line of symmetry, so it meets that side at a right angle and is the triangle’s height. It is a radius followed by an apothem: 12 + 6 = 18. Then \displaystyle \frac{1}{2}\left(12\sqrt{3}\right)(18) = 108\sqrt{3}.

Example 5

An octagon, with the tangent ratio

Find the area of a regular octagon with sides of 10 cm.

Solution

The central angle is \displaystyle \frac{360^\circ}{8} = 45^\circ, so the half slice has an angle of 22.5^\circ at the center. Its leg opposite that angle is half a side, 5, and its leg next to the angle is the apothem.

Figure 3Half of one slice of the octagon. The leg opposite the 22.5^\circ angle is 5, half a side, and the leg next to it is the apothem.

\begin{aligned} \tan 22.5^\circ &= \frac{5}{a} \\ a &= \frac{5}{\tan 22.5^\circ} \approx \frac{5}{0.4142} \approx 12.07 \end{aligned}

The perimeter is 8 \cdot 10 = 80, so

A = \frac{1}{2}(12.07)(80) \approx 482.8

The area is about 482.8\ \text{cm}^2.

Check without trigonometry. Each interior angle of a regular octagon is \displaystyle \frac{(8 - 2) \cdot 180^\circ}{8} = 135^\circ (interior angles of regular polygons). Extend four alternate sides until they meet: they make a square, and the octagon is that square with its four corners cut off.

Figure 4The octagon is a square with its four corners cut off. Each corner is an isosceles right triangle whose hypotenuse is a side of the octagon.

At each end of a cut, the angle between the cut and the square’s side is 180^\circ - 135^\circ = 45^\circ, so each corner is an isosceles right triangle with hypotenuse 10. Its legs are \displaystyle \frac{10}{\sqrt{2}} = 5\sqrt{2} (special right triangles). The square’s side is one leg, one octagon side and another leg, \displaystyle 10 + 10\sqrt{2}, so

\begin{aligned} A &= \left(10 + 10\sqrt{2}\right)^2 - 4 \cdot \frac{1}{2}\left(5\sqrt{2}\right)^2 \\ &= \left(300 + 200\sqrt{2}\right) - 100 \\ &= 200 + 200\sqrt{2} \approx 482.84 \end{aligned}

The two methods agree. The square’s side is also the distance straight across the octagon from one side to the opposite side, which is two apothems, so \displaystyle 2a = 10 + 10\sqrt{2} and \displaystyle a = 5 + 5\sqrt{2} exactly, about 12.07, as the tangent gave.

Example 6

A twelve-sided polygon inside a circle

A regular twelve-sided polygon has its vertices on a circle of radius 10 cm. Find its area, and compare it with the area of the circle.

Solution

The central angle is \displaystyle \frac{360^\circ}{12} = 30^\circ, so the half slice has 15^\circ at the center and hypotenuse 10. With cosine and sine, a = 10\cos 15^\circ \approx 9.659 and \displaystyle \frac{s}{2} = 10\sin 15^\circ, so s \approx 5.176 and P \approx 62.12. Then

A \approx \frac{1}{2}(9.659)(62.12) \approx 300.0

A number that round invites a second look, and a different cut shows it is exactly 300. Take one slice, a triangle with two sides 10 and an angle of 30^\circ between them, and use one of the radii as its base instead of the polygon’s side. Its height is then the perpendicular from the other vertex down to that radius.

Figure 5The twelve-sided polygon misses only the slivers. One slice is shown with its height measured to a radius rather than to a side.

That height is a leg of a right triangle with hypotenuse 10 and 30^\circ at the center: half of an equilateral triangle again, so the leg opposite 30^\circ is 5. The slice has area \displaystyle \frac{1}{2}(10)(5) = 25, and twelve slices make 12 \cdot 25 = 300. The area is exactly 300\ \text{cm}^2.

The circle has area \pi \cdot 10^2 = 100\pi \approx 314.16\ \text{cm}^2. The polygon covers \displaystyle \frac{300}{100\pi} = \frac{3}{\pi} \approx 0.955 of it, about 95.5\%, and what it misses is the twelve thin slivers between each side and its arc.

With more sides the slivers get thinner. The apothem creeps up toward the radius r and the perimeter toward the Circumference 2\pi r, so \displaystyle \frac{1}{2}aP creeps toward \displaystyle \frac{1}{2} \cdot r \cdot 2\pi r = \pi r^2, the area of the circle.

Boundaries

The argument needs a point equally far from every side. It used two facts: the triangles from that point fill the polygon without overlapping, and they all have the same height. A regular polygon always has such a point, its center. A rectangle 2 by 8 does not: its center is 1 from two sides and 4 from the other two, and no single a makes \displaystyle \frac{1}{2}aP right (exercise 20). Cutting from the center still works there, adding triangles of different heights. Some polygons that are not regular do have such a point, every triangle (its incenter) and every rhombus among them, and for those \displaystyle A = \frac{1}{2}aP holds with a that distance (exercise 21).

The apothem is not the radius. The apothem runs from the center to the midpoint of a side, the radius to a vertex. In the half slice the radius is the hypotenuse and the apothem a leg, so r > a always, and putting r into \displaystyle \frac{1}{2}aP gives too large an area.

Same units, square units out. a and P must be in the same unit of length, and the area comes out in that unit squared.

Rounded inputs give rounded areas. An apothem measured or computed to two decimal places gives an area that is only approximate. Keep exact values, such as \displaystyle 3\sqrt{3}, or the full calculator value until the last line, and round once.

The shortcuts belong to their shapes. \displaystyle a = \frac{s}{2} is for the square, \displaystyle a = \frac{\sqrt{3}}{2}s for the hexagon and \displaystyle a = \frac{r}{2} for the equilateral triangle. For other n the angle at the center is not 30^\circ, 45^\circ or 60^\circ, and the tangent or the Pythagorean theorem has to do the work.

Where this goes next

Every regular polygon with n sides is the same shape at a different size, so its apothem and its perimeter are both fixed multiples of its side, and its area is a fixed multiple of s^2. For the hexagon that multiple is \displaystyle \frac{3\sqrt{3}}{2} (exercise 19). Tripling the side multiplies the area by 9, which is how area responds to a scale factor.

A prism or pyramid whose base is a regular polygon needs this area first: the volume of a prism and of a pyramid both start from the area of the base. The faces of a regular pyramid repeat the pattern one dimension up: the lateral surface is half the slant height times the perimeter of the base, with the slant height in the place the apothem holds here.

Letting the number of sides grow, as at the end of the last example, is one road to the area of a circle; another cuts the circle into thin wedges, and Limits make “closer and closer” exact. And for a polygon with no center at all, cutting into triangles and rectangles is the same idea without the symmetry.

Exercises

Area from the apothem and the side

Find the area of each regular polygon.

A regular heptagon (seven sides) with sides of 10 cm and an apothem of 10.4 cm.
A regular octagon with sides of 4 m and an apothem of 4.83 m.
A regular decagon (ten sides) with a perimeter of 50 cm and an apothem of 7.69 cm.
A regular nonagon (nine sides) with sides of 6 cm and an apothem of 8.24 cm.

Finding the apothem first

Find each area exactly, and to two decimal places where it is not a whole number.

A square with sides of 16 cm, using \displaystyle \frac{1}{2}aP. Compare with 16^2.
A regular hexagon with sides of 14 cm.
An equilateral triangle with sides of 12 cm.
A regular hexagon whose apothem is 6 cm.

Working backwards

A regular pentagon has an area of 172\ \text{cm}^2 and an apothem of 6.9 cm. Find its perimeter and its side, each to one decimal place.
A regular octagon has a perimeter of 64 cm and an area of 309\ \text{cm}^2. Find its apothem to two decimal places.
A regular hexagon has an area of \displaystyle 24\sqrt{3}\ \text{cm}^2. Find its side.

With the tangent, sine and cosine

Find the apothem and the area of a regular pentagon with sides of 12 cm, to two and one decimal places.
A regular decagon has its vertices on a circle of radius 8 cm. Find its apothem, its side and its area.

Problems

The floor of a gazebo is a regular hexagon with sides of 3 m. Find its area.
A road sign is a regular octagon measuring 60 cm from one side straight across to the opposite side. Find its side, exactly and to two decimal places, and its area to the nearest square centimeter.
A floor of 12\ \text{m}^2 is to be covered with regular hexagonal tiles with sides of 14 cm. Ignoring gaps and the cutting of tiles at the edges, how many tiles are needed?
Two regular hexagons have sides of 4 cm and 12 cm. How many times the area of the smaller is the area of the larger?

Reasoning

Explain why the triangles made by joining the center of a regular polygon to its vertices are all congruent, and why each one’s height on the polygon’s side is the apothem.
Show that a regular hexagon of side s has area \displaystyle \frac{3\sqrt{3}}{2}s^2.
A rectangle is 2 cm by 8 cm. Its center is 1 cm from the long sides and 4 cm from the short sides. Cut it from the center into four triangles and find its area that way. Then explain why \displaystyle \frac{1}{2}aP gives a wrong answer with a = 1 and with a = 4.
A rhombus has diagonals of 6 cm and 8 cm. Show that its center is the same distance, \displaystyle \frac{12}{5} cm, from all four sides, and that \displaystyle \frac{1}{2}aP with that distance gives its area, even though the rhombus is not regular.
Regular polygons with more and more sides are drawn with their vertices on a circle of radius r. Explain why each one’s area is less than \pi r^2, and why the areas get closer to \pi r^2 as the number of sides grows.

Answers

364\ \text{cm}^2. P = 7 \cdot 10 = 70, and \displaystyle \frac{1}{2}(10.4)(70) = 364.
77.28\ \text{m}^2. P = 8 \cdot 4 = 32, and \displaystyle \frac{1}{2}(4.83)(32) = 77.28.
192.25\ \text{cm}^2. The perimeter is given: \displaystyle \frac{1}{2}(7.69)(50) = 192.25.
222.48\ \text{cm}^2. P = 9 \cdot 6 = 54, and \displaystyle \frac{1}{2}(8.24)(54) = 222.48.
256\ \text{cm}^2. The apothem of a square is half its side, a = 8, and P = 64: \displaystyle \frac{1}{2}(8)(64) = 256 = 16^2.
\displaystyle 294\sqrt{3} \approx 509.22\ \text{cm}^2. The slices are equilateral with side 14, so \displaystyle a = \sqrt{14^2 - 7^2} = \sqrt{147} = 7\sqrt{3}. With P = 84, \displaystyle \frac{1}{2}\left(7\sqrt{3}\right)(84) = 294\sqrt{3}.
\displaystyle 36\sqrt{3} \approx 62.35\ \text{cm}^2. The half slice is half an equilateral triangle, so r = 2a, and (2a)^2 = a^2 + 6^2 gives 3a^2 = 36, \displaystyle a = 2\sqrt{3}. With P = 36, \displaystyle \frac{1}{2}\left(2\sqrt{3}\right)(36) = 36\sqrt{3}.
\displaystyle 72\sqrt{3} \approx 124.71\ \text{cm}^2. The apothem is the height of an equilateral slice, \displaystyle \frac{\sqrt{3}}{2}s = 6, so \displaystyle s = \frac{12}{\sqrt{3}} = 4\sqrt{3} and \displaystyle P = 24\sqrt{3}. Then \displaystyle \frac{1}{2}(6)\left(24\sqrt{3}\right) = 72\sqrt{3}.
P \approx 49.9 cm and s \approx 10.0 cm. \displaystyle P = \frac{2A}{a} = \frac{344}{6.9} \approx 49.86, and \displaystyle s = \frac{P}{5} \approx 9.97.
a \approx 9.66 cm. \displaystyle a = \frac{2A}{P} = \frac{618}{64} = 9.65625.
s = 4 cm. With \displaystyle a = \frac{\sqrt{3}}{2}s and P = 6s, the area is \displaystyle \frac{1}{2} \cdot \frac{\sqrt{3}}{2}s \cdot 6s = \frac{3\sqrt{3}}{2}s^2. Setting this equal to \displaystyle 24\sqrt{3} gives s^2 = 16, and a side is positive.
a \approx 8.26 cm and A \approx 247.7\ \text{cm}^2. The half slice has 36^\circ at the center and opposite leg 6, so \displaystyle a = \frac{6}{\tan 36^\circ} \approx 8.258. With P = 60, A = 30a \approx 247.7.
a \approx 7.61 cm, s \approx 4.94 cm and A \approx 188.1\ \text{cm}^2. The half slice has 18^\circ at the center and hypotenuse 8: a = 8\cos 18^\circ \approx 7.608 and s = 16\sin 18^\circ \approx 4.944. With P = 10s \approx 49.44, \displaystyle \frac{1}{2}(7.608)(49.44) \approx 188.1.
\displaystyle \frac{27\sqrt{3}}{2} \approx 23.38\ \text{m}^2. \displaystyle a = \frac{\sqrt{3}}{2} \cdot 3 = \frac{3\sqrt{3}}{2} and P = 18, so \displaystyle A = \frac{1}{2} \cdot \frac{3\sqrt{3}}{2} \cdot 18.
\displaystyle s = 60\left(\sqrt{2} - 1\right) \approx 24.85 cm, and the area is about 2982\ \text{cm}^2. As in Example 5, the distance across is one side plus two legs of \displaystyle \frac{s}{\sqrt{2}}, which is \displaystyle s\left(1 + \sqrt{2}\right) = 60, so \displaystyle s = \frac{60}{1 + \sqrt{2}} = 60\left(\sqrt{2} - 1\right). The apothem is half the distance across, 30, and P = 8s, so \displaystyle A = \frac{1}{2}(30)(8s) = 7200\left(\sqrt{2} - 1\right).
At least 236 tiles. One tile has area \displaystyle 294\sqrt{3} \approx 509.22\ \text{cm}^2 (exercise 6), and 12\ \text{m}^2 = 120\,000\ \text{cm}^2 since 1\ \text{m}^2 = 100 \cdot 100\ \text{cm}^2. Then \displaystyle \frac{120\,000}{509.22} \approx 235.7, and a part tile still needs a whole one.
9 times. The apothem and the perimeter are both tripled, so \displaystyle \frac{1}{2}aP is multiplied by 3 \cdot 3. Equally, \displaystyle \frac{3\sqrt{3}}{2}(12)^2 \div \frac{3\sqrt{3}}{2}(4)^2 = \frac{144}{16}.
Each triangle has two sides from the center to a vertex, all equal because the center is equally far from every vertex, and a third side that is a side of the polygon, all equal because the polygon is regular. So any two of the triangles are congruent by side-side-side. The height on the polygon’s side is the perpendicular distance from the opposite vertex, the center, to that side, and that perpendicular distance is the definition of the apothem.
The six slices are equilateral with side s, so the apothem is \displaystyle \sqrt{s^2 - \left(\frac{s}{2}\right)^2} = \frac{\sqrt{3}}{2}s, and P = 6s. Then \displaystyle A = \frac{1}{2} \cdot \frac{\sqrt{3}}{2}s \cdot 6s = \frac{3\sqrt{3}}{2}s^2.
The area is 16\ \text{cm}^2. The two triangles on the 8 cm sides have height 1, area \displaystyle \frac{1}{2}(8)(1) = 4 each; the two on the 2 cm sides have height 4, area \displaystyle \frac{1}{2}(2)(4) = 4 each; 4 \cdot 4 = 16 = 2 \cdot 8. With P = 20, a = 1 gives \displaystyle \frac{1}{2}(1)(20) = 10 and a = 4 gives 40. The formula assumes every triangle has the same height, and here two heights differ, so no single a is right.
The diagonals of a rhombus bisect each other at right angles, so they cut it into four congruent right triangles with legs 3 and 4, and hypotenuse \displaystyle \sqrt{3^2 + 4^2} = 5, the side. Each has area \displaystyle \frac{1}{2}(3)(4) = 6; its height from the center onto the side satisfies \displaystyle \frac{1}{2}(5)h = 6, so \displaystyle h = \frac{12}{5}, the same for all four by congruence. The area is 4 \cdot 6 = 24, and \displaystyle \frac{1}{2} \cdot \frac{12}{5} \cdot 20 = 24 as well. The argument only needed equal heights, and the rhombus has them.
The vertices are on the circle, so each side is a chord and the polygon lies inside the circle, leaving a sliver between each side and its arc: its area is less than \pi r^2. In the half slice, the angle at the center is \displaystyle \frac{180^\circ}{n}, which shrinks as n grows, so the leg a gets closer to the hypotenuse r and the slivers get thinner. The perimeter gets closer to the circumference 2\pi r, so \displaystyle \frac{1}{2}aP gets closer to \displaystyle \frac{1}{2} \cdot r \cdot 2\pi r = \pi r^2.