Area of a Polygon Circumscribed About a Circle
The area of a polygon circumscribed about a circle can be calculated using a remarkably simple formula: multiply the polygon's semiperimeter (p) by the radius (r) of its inscribed circle, also known as the incircle. $$ A = p \cdot r $$
The semiperimeter is simply half the perimeter (P) of the polygon:
$$ p = \frac{P}{2} $$
The same formula can also be written in terms of the full perimeter:
$$ A = \frac{P \cdot r}{2} $$
Since P = 2p, substituting this expression into the formula immediately gives:
$$ A = \frac{2p \cdot r}{2} $$
$$ A = p \cdot r $$
This relationship provides a quick and elegant way to find the area of any tangential polygon once its semiperimeter and inradius are known.
Note: In a regular polygon, the radius of the inscribed circle is commonly called the apothem. In more general settings, it is referred to as the inradius.
Why Does the Formula Work?
To understand where the formula comes from, consider a polygon circumscribed about a circle.
As an example, let's look at a hexagon.
The hexagon can be divided into six triangles. Each triangle has a side of the polygon as its base, while the radius of the inscribed circle serves as its height.
The area of each triangle is:
$$ \frac{l \cdot r}{2} $$
where l represents the length of the corresponding side.
Because the polygon is made up of these triangles, its total area is simply the sum of their areas.
$$ A = \frac{ \overline{AB} \cdot r }{2} + \frac{ \overline{BC} \cdot r }{2} + \frac{ \overline{CD} \cdot r }{2} + \frac{ \overline{DE} \cdot r }{2} + \frac{ \overline{EF} \cdot r }{2} + \frac{ \overline{FA} \cdot r }{2} $$
Factoring out the common term r/2 gives:
$$ A = \frac{ ( \overline{AB} + \overline{BC} + \overline{CD} + \overline{DE} + \overline{EF} + \overline{FA} ) \cdot r }{2} $$
The expression inside the parentheses is the perimeter P of the polygon, so:
$$ A = \frac{P \cdot r}{2} $$
Replacing P with 2p produces:
$$ A = \frac{2p \cdot r}{2} $$
and therefore:
$$ A = p \cdot r $$
This is exactly the formula we wanted to prove.
Geometric Interpretation: Another way to visualize the result is to imagine rearranging all the triangular regions into a single triangle whose base is the perimeter P and whose height is the inradius r. The area of this new triangle is equal to the area of the original polygon, leading directly to the formula.
Useful Formulas
The area formula can be rearranged to find other quantities when the remaining values are known.
- Finding the Inradius
If you know the area and the semiperimeter: $$ r = \frac{A}{p} $$ - Finding the Semiperimeter
If you know the area and the inradius: $$ p = \frac{A}{r} $$
These relationships are often useful when solving geometry problems involving tangential polygons and inscribed circles.
