Area of a Circle

The Circle

Before knowing about the area of a circle let us know about what a circle is. A circle is a simple closed shape. It is the set of all points in a plane that are at a given distance from a given point, the centre; equivalently it is the curve traced out by a point that moves so that its distance from a given point is constant. The distance between any of the points and the centre is called the radius. The constant π is traditionally defined as the ratio of a circle’s circumference to its diameter. The fact that this is well-defined (i.e., it is independent of the circle chosen) requires proof and the proof is given in Euclid’s elements actually involves area rather than circumference!

In particular, Euclid proved that the areas of two circles are proportional as the ratio of the squares of their diameters, and so it follows that the quantity A/r², where r is the radius of the circle is independent of the circle. If we call this ratio π (which the Greeks did not, by the way; that came much later), then we get the familiar formula A=π r².

A circle (black) which is measured by its circumference (C), diameter (D) in cyan, and radius (R) in red; its centre (O) is in magenta.

Historically speaking, the Area of a Circle (or more accurately, the area enclosed by a circle) has captured the fascination of mathematicians from times as early as that of Archimedes. In fact, the formula for the area of the circle has been derived earliest by Archimedes himself in his book On the Measurements of the Circle. The formula for area of a circle is as follows:

Proofs

1. Proof by Rearrangement

We can use the regular polygons “fitted” within a circle. Suppose we inscribe a hexagon. Cut the hexagon into six triangles by splitting it from the centre. Then, the two opposite triangles both touch two common diameters; slide them along one so the radial edges are adjacent. They now form a parallelogram, with the hexagon sides making two opposite edges, one of which is the base, s. Two radial edges form slanted sides, and the height, h is equal to its apothem. In fact, we can assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase to eight sides and so on. For a polygon with 2n sides, the parallelogram will have a base of length ns, and a height h. As the number of sides increases, the length of the parallelogram base approaches half the circle circumference, and its height approaches the circle radius. In the limit, the parallelogram becomes a rectangle with width πr and height r.

             (h is the apothem here)

Hence, the area of our rectangle becomes π r2 !

Example:

  1. Cut a circle into equal sectors (12 in this example)
  2. Cut one sector in half and rearrange.

  3. This would resemble a rectangle.
  4. So, what are the (approximate) height and width of the rectangle?The height is the circle’s radius: just look at sectors 1 and 13 above. When they were in the circle they were “radius” high.The width (actually one “bumpy” edge) is half of the curved parts around the circle … in other words it is about half the circumference of the circle.We know that:

    Circumference = 2 × π × radius

    And so the width is about:

    Half the Circumference = π × radius

    And so we have (approximately):

π€ × radius

Now we just multiply the width by the height to find the area of the rectangle:

Area = (π × radius) × (radius)

π × radius2

As we would increase the sectors, we would approach to a more fine or a “less bumpy” rectangle.

 

2. Triangle proof

Archimedes used it to calculate the area of a circle with a different formulation than Euclid’s assertion that it is proportional to r². His theorem is that the area of a circle equals the area of a triangle whose height is the radius r and whose base is the circumference of the circle. In symbols: A = (½)·r·C, which, when put together with the other area formula yields C = 2πr. This is a rather remarkable statement since nowhere does Euclid mention the length of any curve except a line segment, and indeed, even defining the length of a curve that isn’t just a line segment is problematic. Really, it requires calculus and the idea that a curve is the limit of straight line segments connecting points on the curve.

Consider unwrapping the concentric circles into straight strips. This will form a right-angled triangle with r as its height and 2πr (being the outer slice of onion) as its base.

Finding the area of this triangle will give the area of the disk

The circle and the triangle are equal in area.

 

3. Proof by Limiting Polygons

For any regular polygon escribing a circle, the limiting shape of the polygon on increasing its length approaches the shape of a circle.

We see that the square which earlier escribed the circle when gets replaced by an octagon, the overall shape of the polygon is closer to that of the circle.

Now, consider the following figure:

The area of triangle AOB is 1/2 ( base × height) = 1/2 (s × r)
This means that the area of the entire octagon is 8 ×( 1/2 (s × r)) = 1/2 r × 8s

Notice that 8s is equal to the perimeter of the octagon.  As stated before, if we increase the number of sides to infinity or a very big number, the resulting n-gon ( The regular polygon which the number of sides is a big number) will almost look like a circle. This means that the perimeter of the octagon will almost be the same as the perimeter of the circle. As a result, the closer the perimeter of the polygon is to the circle, the closer the area of the polygon is to the area of the circle.

It is reasonable then to replace 8s by 2 × π × r, which is the perimeter of the circle, to calculate the area of the polygon or the circle when the number of sides is very big

Doing so we get:

Area of circle or polygon equal = 1/2 r × 2 × π × r = π × r2  !

The area and perimeter ratios are shown and as n, the number of sides, gets larger we get a better approximation to π=3.14159…

 

Also read, area of a rectangle article.

For more such articles follow us at Toppr Bytes Good Luck!

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Vibhu
Vibhu

Physics sophomore at IIT KGP. www.vibhuist.wordpress.com

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