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

Example:

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


circum-diameter

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} = 16.33cm

Comments

comments