Problem 1: A 6 sided regular polygon (hexagon) is inscribed in a circle of radius 10 cm, find the length of one side of the hexagon.
Solution to Problem 1:

Angle AOB is given by
angle (AOB) = 360^{o} / 6 = 60^{o}

Since OA = OB = 10 cm, triangle OAB is isosceles which gives
angle (OAB) = angle (OBA)

So all three angles of the triangle are equal and therefore it is an equilateral triangle. Hence
AB = OA = OB = 10 cm.
Problem 2: A circle of radius 6 cm is inscribed in a 5 sided regular polygon (pentagon), find the length of one side of the pentagon.(approximate your answer to two decimal places).
Solution to Problem 2:

Let t be the size of angle AOB, hence
t = 360^{o} / 5 = 72^{o}

The polygon is regular and OA = OB. Let M be the midpoint of AB so that OM is perpendicular to AB. OM is the radius of the inscribed circle and is equal to 6 cm. Right angle trigonometry gives
tan(t / 2) = MB / OM

The side of the pentagon is twice MB, hence
side of pentagon = 2 OM tan(t / 2) = 8.7 cm (answer rounded to two decimal places)
Problem 3: Find the area of a dodecagon of side 6 mm. (approximate your answer to one decimal place).
Solution to Problem 3:

A dodecagon is a regular polygon with 12 sides and the central angle t opposite one side of the polygon is given by.
t = 360^{o} / 12 = 30^{o}

We now use the formula for the area when the side of the regular polygon is known
Area = (1 / 4) n x^{2} cot (180^{o} / n)

Set n = 12 and x = 6 mm
area = (1 / 4) (12) (6 mm)^{2} cot (180^{o} / 12)
= 403.1 mm^{2} (approximated to 1 decimal place).
Problem 4:Show that if the number of sides n of a polygon inscribed inside a circle of radius R, is very large then the area of the polygon may be approximated by the area of the circumscribed circle with radius R. (HINT: If angle x is very small and is in radians, then sin x may be approximated by x).
Solution to Problem 4:

The area of a regular polygon with n sides may be given in terms of R by
area = (1/2) n R ^{2} sin (2 pi / n)

If n is large, then 2 Pi /n is very small and sin (2 pi/n) may be approximated by 2 pi / n so that the area may be approximated by
area = (1/2) n R ^{2} (2 pi / n)
= pi R ^{2}

which is the area of the circle.
For more on the above question, see the interactive tutorial in regular polygons.
More references to triangles and geometry.
Geometry Tutorials, Problems and Interactive Applets.
