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Areas of Regular Polygons: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Angles at the center

A convex regular polygon has 20 sides. Its center is O, B and C are adjacent vertices, and OM is perpendicular to BC at M.

Find the central angle BOC and the half-central angle BOM. Explain why MB is half a side.

One right half of a center triangle; the polygon count is supplied in the text.

Text description of this figure

A schematic right half-triangle OMB. OM meets MB at a right angle, OB is the radius, OM is the apothem, and MB is half a side. The full polygon has 20 sides as specified in the statement.
1. Hint 1
  The congruent center triangles surround O once.
2. Hint 2
  The two radii make OBC isosceles. Compare its right halves at M.
Answer

$\begin{aligned}\angle BOC&=18^\circ\\\angle BOM&=9^\circ.\end{aligned}$
MB is half of BC because the two right halves are congruent.

Full solution

There are 20 equal angles at O, filling 360 degrees.
$\angle BOC=360^\circ\div20.$
Thus angle BOC is 18 degrees.

The equal radii OB and OC are hypotenuses of right triangles OMB and OMC, which share leg OM.
Hypotenuse-leg congruence makes their angles at O equal and their base parts MB and MC equal.

Halving the central angle gives
$\angle BOM=9^\circ.$
The equal base parts make M the midpoint, so MB is half of BC.

Answer

$\begin{aligned}\angle BOC&=18^\circ\\\angle BOM&=9^\circ.\end{aligned}$
MB is half of BC because the two right halves are congruent.

Key idea

The perpendicular in an isosceles center triangle bisects both the side and the central angle.
Problem 2 A square from its radius

A square has center O and radius $6\sqrt2$ cm. Find its apothem, side length, and area.

Use the given hypotenuse to recover both legs.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius 6√2 cm. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  Use the right half of a center triangle.
2. Hint 2
  Its half-central angle is 45 degrees, so the legs are equal and each is the hypotenuse divided by the square root of 2.
Answer

$\begin{aligned}a&=6\text{ cm}\\s&=12\text{ cm}\\A&=144\text{ cm}^2.\end{aligned}$

Full solution

The half-central angle is 45 degrees.
In its 45-45-90 triangle, each leg is $6\sqrt2/\sqrt2$ , or 6 cm.
The apothem is one leg, and the other leg is half a side.

Doubling the half-side gives a side of 12 cm.
The perimeter is 48 cm.

Use the apothem as the perpendicular height:
$A=\frac12(48)(6).$
This gives an area of 144 square cm.
Squaring the 12 cm side confirms the result.

Use the given hypotenuse to recover both legs.

Answer

$\begin{aligned}a&=6\text{ cm}\\s&=12\text{ cm}\\A&=144\text{ cm}^2.\end{aligned}$

Key idea

The radius of a square is the hypotenuse of its 45-45-90 center half-triangle.
Problem 3 A hexagon from its perimeter

A convex regular hexagon has perimeter 66 cm. Find its side length, apothem, and exact area.

Find the missing perpendicular height from the center triangle.

Text description of this figure

Schematic convex regular 6-gon with center O. Segments from O to every vertex show center triangles. The bottom side is labeled side s. A dashed perpendicular from O meets the midpoint M of that side, is labeled a, and has a right-angle mark. The radius to the right endpoint is labeled r. No computed answers are labeled.
1. Hint 1
  Find one side before looking for the height.
2. Hint 2
  Each center triangle is equilateral; its perpendicular bisector creates a 30-60-90 triangle.
Answer

$\begin{aligned}s&=11\text{ cm}\\a&=\frac{11\sqrt3}{2}\text{ cm}\\A&=\frac{363\sqrt3}{2}\text{ cm}^2.\end{aligned}$

Full solution

The six equal sides share the perimeter.
$s=66\div6.$
Thus each side is 11 cm.

A hexagon's center triangles are equilateral, so each radius is 11 cm.
The half-side is 11/2 cm.
The 30-60-90 ratio gives
$a=\frac{11\sqrt3}{2}\text{ cm}.$

The area uses the full perimeter:
$A=\frac12(66)\frac{11\sqrt3}{2}.$
Therefore
$A=\frac{363\sqrt3}{2}\text{ cm}^2.$

Answer

$\begin{aligned}s&=11\text{ cm}\\a&=\frac{11\sqrt3}{2}\text{ cm}\\A&=\frac{363\sqrt3}{2}\text{ cm}^2.\end{aligned}$

Key idea

A hexagon perimeter determines a side, and its equilateral center triangles determine the height.
Problem 4 Recover a center distance

An equilateral triangle has area $27\sqrt3$ square cm and perimeter $18\sqrt3$ cm. Find its apothem. Then find its side length and check the apothem with a special right triangle.

The unknown apothem is adjacent to the half-central angle.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius r. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  Area and perimeter determine the common perpendicular height.
2. Hint 2
  Solve the perimeter-apothem equation for the apothem, then divide the perimeter by three.
Answer

$\begin{aligned}a&=3\text{ cm}\\s&=6\sqrt3\text{ cm}.\end{aligned}$
The 60-degree half-triangle has a half-side of $3\sqrt3$ cm, compatible with this apothem.

Full solution

Double the area and divide by the positive perimeter:
$a=\frac{54\sqrt3}{18\sqrt3}.$
The apothem is 3 cm.

Three equal sides share the perimeter, giving side length $6\sqrt3$ cm.
Half a side is $3\sqrt3$ cm.

The half-central angle is 60 degrees.
The half-side is the longer leg and is the apothem times the square root of 3.
Thus the side-derived apothem is also 3 cm.

Answer

$\begin{aligned}a&=3\text{ cm}\\s&=6\sqrt3\text{ cm}.\end{aligned}$
The 60-degree half-triangle has a half-side of $3\sqrt3$ cm, compatible with this apothem.

Key idea

An inverse area calculation should agree with the geometry of the right half-triangle.
Problem 5 A pentagon calculation

A convex regular pentagon has side length 9 cm. Find an exact trigonometric expression for its apothem and for its area. Then give the area to the nearest tenth of a square centimeter. Use degree mode.

The half-side and apothem are the tangent pair.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius r. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  Find the half-central angle and the half-side.
2. Hint 2
  Tangent compares the half-side with the apothem; retain its exact expression when finding area.
Answer

$a=\frac9{2\tan36^\circ}\text{ cm}.$ $A=\frac{405}{4\tan36^\circ}\text{ cm}^2.$
The area is approximately 139.4 square cm.

Full solution

The half-central angle is 36 degrees and half a side is 4.5 cm.
Tangent gives
$a=\frac{4.5}{\tan36^\circ}.$

The perimeter is 45 cm.
Hence
$A=\frac12(45)\frac{4.5}{\tan36^\circ}.$
Grouping the constants gives
$A=\frac{405}{4\tan36^\circ}\text{ cm}^2.$

Evaluate that expression in degree mode without first rounding the apothem.
The area is approximately 139.4 square cm.

The half-side and apothem are the tangent pair.

Answer

$a=\frac9{2\tan36^\circ}\text{ cm}.$ $A=\frac{405}{4\tan36^\circ}\text{ cm}^2.$
The area is approximately 139.4 square cm.

Key idea

Use the half-central angle, and delay rounding until after the area calculation.
Problem 6 A flat octagonal panel

A flat panel is modeled as a convex regular octagon of radius 7 cm. Only one face is to be coated. Find the area to be coated, to the nearest tenth of a square centimeter. Ignore the thickness of the panel and use degree mode.

The schematic gives the one-face model its side and height relationships.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius 7 cm. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  The given center-to-vertex distance is a radius, so first find the perpendicular height and a side.
2. Hint 2
  Cosine finds the apothem; sine finds half a side. Use all eight sides for the perimeter.
Answer

Approximately $138.6\text{ cm}^2$ .

Full solution

The half-central angle is 22.5 degrees.
Cosine gives
$a=7\cos22.5^\circ.$
Sine gives
$s=14\sin22.5^\circ.$
Both lengths are in centimeters.

The eight side lengths total $8s$ .
Use the unrounded side and apothem:
$A=\frac12(8s)a.$

Substituting both trigonometric expressions gives approximately 138.6 square cm.
The model coats one flat face, so the result is not doubled.

The schematic gives the one-face model its side and height relationships.

Answer

Approximately $138.6\text{ cm}^2$ .

Key idea

A center-to-vertex measurement supplies a hypotenuse, from which both triangle legs are needed.
Problem 7 Compare the two center distances

For every nondegenerate convex regular polygon, let s be the positive side length, a the apothem, and r the radius. Prove that the radius is longer than the apothem. State an equation connecting these three lengths and explain where positivity matters.

The equation uses the radius as hypotenuse.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius r. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  Work in one right half-triangle.
2. Hint 2
  Apply the Pythagorean theorem to the hypotenuse r and legs a and half a side.
Answer

$r^2=a^2+(s/2)^2.$
For positive side length,
$r>a.$

Full solution

For a numerical instance, a square with apothem 11 units has half-side 11 units and radius $11\sqrt2$ units.
Pythagoras adds two positive leg-squares, so that radius is longer than either leg.

For the general polygon, the perpendicular to a side bisects that side.
The right half-triangle therefore has legs a and s/2, and hypotenuse r.
The Pythagorean theorem gives
$r^2=a^2+(s/2)^2.$

The side length is positive, so $(s/2)^2$ is positive.
Thus
$r^2>a^2.$
Both distances are positive, so taking positive square roots preserves their order and gives
$r>a.$
If the side length were zero, the strict-inequality step would fail; that degenerate case is excluded.

Answer

$r^2=a^2+(s/2)^2.$
For positive side length,
$r>a.$

Key idea

The radius is longer because a nonzero half-side contributes a positive square.
Problem 8 Test two center distances

Determine whether a convex regular polygon can have radius exactly 17 cm and apothem exactly 15 cm. Find the side length that these distances would require, then check whether the required side count is possible. Use degree mode.

Test the valid right triangle before deciding whether copies can close into a regular polygon.

Text description of this figure

A schematic right triangle OMB with hypotenuse OB labeled radius 17 cm, perpendicular leg OM labeled apothem 15 cm, and leg MB labeled half-side. It depicts a valid right triangle only; no regular polygon with these distances is asserted.
1. Hint 1
  The radius and apothem first determine the half-side in a right triangle.
2. Hint 2
  Use the half-side and apothem to find the acute half-central angle with inverse tangent.
3. Hint 3
  Compare the resulting side count with integers; do not round it into an integer.
Answer

The required side length would be 16 cm. No convex regular polygon has these exact distances: its required side count lies strictly between 6 and 7.

Full solution

A right half-triangle with hypotenuse 17 cm and adjacent leg 15 cm would have opposite leg t satisfying
$t^2=17^2-15^2.$
Hence t is 8 cm and the required full side would be 16 cm.
This right triangle exists; whether copies fit a regular polygon still needs checking.

Its acute half-central angle would be
$\theta=\tan^{-1}(8/15).$
In degree mode this is approximately 28.0725 degrees.

A regular polygon would then need
$n=180^\circ/\theta.$
This gives approximately 6.41197 sides, strictly between 6 and 7.
Since the exact count must be an integer, these distances cannot be the radius and apothem of a convex regular polygon.

Rounding the required side count to 6 or 7 would change the half-central angle and therefore change at least one exact distance.
Fitting a right triangle is necessary but does not alone ensure that its copies close into a regular polygon.

Answer

The required side length would be 16 cm. No convex regular polygon has these exact distances: its required side count lies strictly between 6 and 7.

Key idea

Compatible right-triangle lengths must also give an integer regular-polygon side count.
Problem 9 Eighteen center triangles

A convex regular 18-gon has radius 13 cm. Find its apothem and perimeter to the nearest tenth of a centimeter, and its area to the nearest tenth of a square centimeter. Use degree mode and keep full calculator precision until each final result.

A right half-triangle relates the radius, side, and apothem.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem a, MB is labeled half-side s/2, and OB is labeled radius 13 cm. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  The half-central angle is smaller than the full central angle.
2. Hint 2
  Find the adjacent leg and opposite leg relative to the half-central angle; double the opposite leg for a side.
Answer

$\begin{aligned}a&\approx12.8\text{ cm}\\P&\approx81.3\text{ cm}\\A&\approx520.2\text{ cm}^2.\end{aligned}$

Full solution

The half-central angle is 10 degrees.
The adjacent-leg calculation gives
$a=13\cos10^\circ.$
This is approximately 12.8 cm.

Twice the opposite leg is a side, so
$s=26\sin10^\circ.$
All 18 sides give
$P=468\sin10^\circ.$
This is approximately 81.3 cm.

Use the unrounded perimeter and apothem:
$A=\frac12 P(13\cos10^\circ).$
This gives approximately 520.2 square cm.

One of the 36 right halves determines the side and apothem.

Answer

$\begin{aligned}a&\approx12.8\text{ cm}\\P&\approx81.3\text{ cm}\\A&\approx520.2\text{ cm}^2.\end{aligned}$

Key idea

Retain full precision in both the side and height when a polygon has many triangles.
Problem 10 Recover the side count

A convex regular polygon has side length 22 cm and apothem $11\sqrt3$ cm. These are exact lengths. Find its number of sides, radius, and exact area. Check that the side count is a valid integer and that the three lengths fit a right half-triangle.

The exact leg ratio determines the side count; the schematic supplies no angle measurement.

Text description of this figure

An enlarged schematic right triangle OMB. OM is perpendicular to MB at M. OM is labeled apothem 11√3 cm, MB is labeled half-side s/2, and OB is labeled radius r. The angle at O is labeled θ. Lengths and angle size must come from the text, not measurement.
1. Hint 1
  The side and apothem determine the tangent of the half-central angle.
2. Hint 2
  Identify the acute angle whose tangent is the half-side divided by the apothem. Then recover the side count.
3. Hint 3
  Use the resulting special right triangle for the radius, and all the sides for the perimeter.
Answer

$\begin{aligned}n&=6\\r&=22\text{ cm}\\A&=726\sqrt3\text{ cm}^2.\end{aligned}$
The side count is an integer at least 3. The lengths satisfy
$22^2=(11\sqrt3)^2+11^2.$

Full solution

Half a side is 11 cm.
For the half-central angle theta,
$\tan\theta=\frac{11}{11\sqrt3}.$
Thus its tangent is $1/\sqrt3$ , so the acute angle is 30 degrees.

The side count is
$n=180\div30.$
Hence n is 6, a valid integer at least 3, and the polygon is a regular hexagon.

The half-triangle is 30-60-90 with shorter leg 11 cm.
Its hypotenuse, the radius, is 22 cm.
For a direct check, the two leg-squares add to 484, which is the square of 22.

Six sides give perimeter 132 cm.
The exact area is
$A=\frac12(132)(11\sqrt3).$
Thus
$A=726\sqrt3\text{ cm}^2.$

The exact leg ratio determines the side count; the schematic supplies no angle measurement.

Answer

$\begin{aligned}n&=6\\r&=22\text{ cm}\\A&=726\sqrt3\text{ cm}^2.\end{aligned}$
The side count is an integer at least 3. The lengths satisfy
$22^2=(11\sqrt3)^2+11^2.$

Key idea

The side-to-apothem ratio determines the half angle, which must give an integer side count.

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