# Relation Between the Circumference and Diameter

It we measure the circumference and diameter of various circles; we will find that the ration of circumference and their corresponding diameters, is always constant. This constant ratio is denoted by 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 circumference is 8 cm. Find the two circumferences.

Solution:
Let $x$ be the radius of one circle.
The radius of the other circle $= (7 - x)$ cm
Now, circumference of circle with radius $x$ cm $= 2\pi x$cm --- (1)
and, circumference of circle with radius $(7 - x)$cm $= 2\pi (7 - x)$ --- (2)
The difference of two circumference = 8 cm --- (3)
From equations (1), (2) and (3) we get
$\therefore$     $2\pi x - 2\pi (7 - x) = 8$
$2\pi x - 14\pi + 2\pi x = 8$
$4\pi x = 44 + 8 = 52$
$\therefore$     $2\pi x = \frac{{52}}{2} = 26$cm
Thus, circumference of the circle with radius $x$cm =26 cm and circumference of second circle, $= 2\pi (7 - x)$
$= 14\pi - 2\pi x$
$= 14 \times \frac{{22}}{7} - 26$
$= 44 - 26 = 18$cm

Example:

What will be the circumference of a circle inscribed an equilateral triangle of side 9 cm.

Solution:

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

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

Now 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 $= 2\pi r$

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