Circle-Triangle Area Equivalence Theorem
A circle's area matches that of a triangle whose base equals the circle's circumference and whose height is the circle's radius. $$ \pi \cdot r^2 = \frac{1}{2} \cdot c \cdot r $$
Put simply, a circle equates to a triangle sharing its circumference as the base and its radius as the height.
To calculate a circle's area, multiply pi (π=3.14...) by the radius squared (r).
$$ A = \pi r^2 $$
Conversely, the circle's circumference \( c \) is twice the product of pi and the radius (r).
$$ c = 2\pi r $$
For a triangle with the circle's circumference (\(c = 2\pi r\)) as its base and the circle's radius (\(h = r\)) as its height, the area \(A_t\) is given by:
$$ A_t = \frac{1}{2} \cdot base \cdot height = \frac{1}{2} \cdot C \cdot r = \frac{1}{2} \cdot 2\pi r \cdot r = \pi r^2 $$
Therefore, the circle and this particular triangle share identical areas.
Note: While this theorem isn't widely mentioned in standard geometry texts, it offers a fascinating link between the areas of a circle and a triangle. It's more an insightful geometric observation than a specifically named theorem, directly arising from the formulas for calculating the areas of circles and triangles.
The Proof
Imagine a circle with circumference c and radius r.
Next, create a regular polygon inscribed within this circle, like a hexagon.
This leads to the conclusion that the area of the inscribed polygon $ A_p $ is less than the circle's area $ A_c $.
$$ A_p < A_c $$
Then, construct a regular polygon circumscribed around the circle, with an equal number of sides.
From this, it's inferred that the area of the circumscribed polygon $ A_{p'} $ is greater than the circle's area $ A_c $.
$$ A_p < A_c < A_{p'} $$
Thus, the circle's area $ A_c $ acts as a boundary between the areas of inscribed and circumscribed regular polygons.
Increasing the sides of these polygons brings their areas $ A_p $ closer to the circle's area $ A_c $.
Extending this idea to infinity suggests a regular polygon with infinite sides having the same area as the circle.
$$ A_c = A_p $$
According to the theorem on circumscribable polygons, any regular polygon is equivalent to a triangle whose base matches the polygon's perimeter and whose height equals the polygon's apothem.
Note: For a regular polygon with infinite sides, the apothem aligns with the radius of the encompassing circle, making the triangle's height equal to the circle's radius.
Thus, the area of the infinite-sided polygon equals that of the corresponding triangle:
$$ A_c = A_p = A_t = \frac{1}{2} \cdot p \cdot a $$
Where $ p $ is the polygon's perimeter and $ a $ is its apothem.
Given that the regular polygon's perimeter equals the circumference $ p = c $ and the apothem equals the circle's radius $ a = r $:
$$ A_c = A_p = A_t = \frac{1}{2} \cdot c \cdot r $$
In conclusion, the area of the circle $ A_c $ matches the area of the triangle whose base is the circumference and whose height is the radius.
$$ A_c = \frac{1}{2} \cdot c \cdot r $$
Quick Check: Remembering the circle's area formula $ A_c = \pi r^2 $ and comparing it $$ A_c = \frac{1}{2} \cdot c \cdot r $$ $$ \pi r^2 = \frac{1}{2} \cdot c \cdot r $$ with the circumference formula $ c = 2 \pi r $ $$ \pi r^2 = \frac{1}{2} \cdot ( 2 \pi r ) \cdot r $$ simplifies to $$ \require{cancel} \pi r^2 = \pi r^2 $$ confirming the theorem's validity.
And so on.