Relation Between the Circumference and Diameter

It we measure the circumference and diameter of various circles, we will find that the ratio of the circumference and their corresponding diameters is always constant. This constant ratio is denoted by the Greek letter \pi .

The value of \pi is 22/7or 3.1415.
\therefore    \frac{{{\text{Circumference}}}}{{{\text{Diameter}}}} = \pi
Or    {\text{Circumference = }}\pi \times {\text{ Diameter}}
= \pi d  Or  2\pi r        (r, being radius)

 

Example:

The sum of two radii of two circles is 7cm and the difference of their circumferences is 8cm. Find the two circumferences.

Solution:

            Let x be the radius of one circle.
The radius of the other circle is = (7 - x) cm
Now, the circumference of the circle with radius x cm = 2\pi xcm --- (1)
and, the circumference of the circle with radius (7 - x)cm = 2\pi (7 - x) --- (2)
The difference of the two circumferences = 8 cm --- (3)

            From equations (1), (2) and (3) we get
\therefore      2\pi x - 2\pi (7 - x) = 8
2\pi x - 14\pi + 2\pi x = 8
4\pi x = 44 + 8 = 52
\therefore      2\pi x = \frac{{52}}{2} = 26cm

Thus, the circumference of the circle with radius xcm =26 cm and the circumference of the second circle, = 2\pi (7 - x)
= 14\pi - 2\pi x
= 14 \times \frac{{22}}{7} - 26
= 44 - 26 = 18cm

 

Example:

Find the circumference of a circle inscribed in an equilateral triangle of side 9 cm.


circum-diameter
 

Solution:

Let O be the centre of the inscribed circle of \Delta ABC

Since the radius of the inscribed circle in a triangle = \frac{{{\text{Area of }}\Delta ABC}}{{{\text{Semi perimeter of }}\Delta ABC}}

Now the area of \Delta ABC = \frac{{\sqrt 3 }}{4}{({\text{one side}})^2} = \frac{{\sqrt 3 }}{4}{(9)^2}

\because Semi perimeter of \Delta ABC = \frac{{27}}{2}

\therefore Radius of inscribed circle = \frac{{\frac{{\sqrt 3 }}{4}{{(9)}^2}}}{{\frac{{27}}{2}}} = \frac{{3\sqrt 3 }}{2}cm
\therefore Circumference of the circle = 2\pi r

= 2 \times \frac{{22}}{7} \times \frac{{3\sqrt 3 }}{2} = 16.33cm