Area of a circle - limit of triangles
Peter Haggstrom
www.gotohaggstrom.com
[email protected]
January 10, 2013
1 Background
As a simple high school exercsise, if you want to prove that the area of a circle of radius
r is π r2 you can approximate the area by n triangles with angle 2π
n and then let n → ∞.
This simple approach to approximation uses a technique which is applicable in several
other contexts.
2 How the approximation works
The diagram below gives you the set up:
1
A
l
B
l
r
h
C
Π
n
Π
n
r
O
2
The area of triangle OAB is simply 12 × l × r and so the area of the double triangle OAC
is l × r. Because angle∠OBA is a right angle, l = r sin( πn ) and h = r cos( πn ).
The area of triangle OAC is thus r sin( πn ) r cos( πn ) =
r2
2
sin( 2π
n )
2
The total circular angle of 2π is thus divided into n equal triangles of area r2 sin( 2π
n ) so
that the area of the circle is approximated by the limit of this sum as n → ∞:
Area of circle = lim n
n→∞
since
sin( 2π
)
n
2π
n
sin( 2π )
2π
r2
sin( ) = lim πr2 2πn = πr2
n→∞
2
n
n
(1)
→ 1 as n → ∞
There are other ways of getting the same result. For instance, consider an elementary
sector with small angle dθ. The area of this sector is approximated by 12 r2 sin(dθ) and
because dθ is small we can approximate
by dθ ( this is because sin x ≈ x for
R 2πsin(dθ)
1 2
small x.) The area of the circle is then 0 2 r dθ = π r2
3