Properties of a Regular Polygon
These are the properties of a regular polygon:
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The sides and interior angles of a regular polygon are all equal.
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The bisectors of the interior angles of a regular polygon meet at its center.
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The perpendiculars drawn from the center of a regular polygon to its sides are all equal.
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The lines joining the center of a regular polygon to its vertices are all equal.
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The center of a regular polygon is the center of both the inscribed and circumscribed circles.
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Straight lines drawn from the center to the vertices of a regular polygon divide it into as many equal isosceles triangles as there are sides in it.
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The angle of a regular polygon of $$n$$ sides $$ = \left( {\frac{{2n – 4}}{n}} \right) \times {90^ \circ }$$.
Detail of the Sides of a Polygon
There is no theoretical limit to the number of sides of a polygon, but only those with twelve or less are commonly seen. The names of polygons which are used most often are as follows:
Number of sides
|
Polygon name
|
$$5$$
|
Pentagon
|
$$6$$
|
Hexagon
|
$$7$$
|
Heptagon
|
$$8$$
|
Octagon
|
$$9$$
|
Nonagon
|
$$10$$
|
Decagon
|
$$ \cdots $$
|
$$ \cdots $$
|
$$ \cdots $$
|
$$ \cdots $$
|
$$n$$
|
$$n – $$gon
|
Example:
The perimeter and area of a regular polygon are respectively equal to those of a square of sides $$a$$. Find the length of the perpendicular from the center of a regular polygon to any of its sides.
Solution:
The perimeter of a regular polygon = perimeter of square $$ = 4a$$
A regular polygon can be divided into congruent triangles having a common vertex at the center of the polygon. The number of these congruent triangles is the same as that of its sides.
$$\therefore $$ the area of one such triangle $$ = \frac{1}{2} \times $$ sides of polygon $$ \times $$ length of perpendicular from the center to any side of polygon.
$$\therefore $$ the area of one such triangle$$ = \frac{1}{2}ah$$, $$h$$ being the length of the perpendicular
$$\therefore $$ Area of polygon = Sum of areas of all such triangles
$$\therefore $$ Area of polygon $$ = \frac{1}{2} \times $$ Perimeter of polygon $$ \times {\text{ }}h$$
$$\therefore $$ Area of polygon $$ = \frac{1}{2} \times 4a \times h = 2ah$$ — (1)
Now, the area of the polygon = Area of square$$ = {a^2}$$ (given) — (2)
$$\therefore $$ from (1) and (2), we have
$$\therefore $$ $$2ah = {a^2} \Rightarrow h = \frac{a}{2}$$
Hence, the length of the perpendicular is $$\frac{a}{2}$$.