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.

Circle and triangle equivalence theorem visual

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.

    Circle illustration

    Next, create a regular polygon inscribed within this circle, like a hexagon.

    Hexagon inscribed in a circle

    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.

    Hexagon circumscribed around a circle

    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 $.

    Polygon with more sides inscribed in a circle

    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.

    Regular polygon and triangle equivalence theorem visualization

    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.

     
     

    Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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