# Area of a Circular Ring or Annulus

Annulus:

A circular ring (annulus) is plane figure bounded by the circumference of two concentric circles of two different radii. The area of a circular ring is found by subtracting the area of the small circle from that of the large circle. An example of an annulus is the area of a washer and the area of a concrete pipe.

If $A$ and $a$, $R$ and $r$ stand for the areas and the radii of two circles and ${A_r}$ for the area of the ring, the

${A_r} = A – a = \pi {R^2} – \pi {r^2} = \pi ({R^2} – {r^2}) = \pi (R – r)(R + r)$  i.e. to find the area of a ring (or annulus), multiply the product of the sum and the difference of the two radii by $\pi$ in the first figure.

Note: Rule holds good even when circles are not concentric as in second figure.

Example:

A path $14$cm wide surrounds a circular lawn with a diameter of $360$cm. Find the area of the path.

Solution:

Given that
Radius of inner circle $= 180$cm
Radius of outer circle $= (180 + 14) = 194$cm
$\therefore$ the area of path $= \pi (R – r)(R + r)$
$= \frac{{22}}{7}(194 + 180)(194 – 180) = 16456$ Square cm

Example:

The areas of two concentric circles are $1386$square cm and $1886.5$ square cm respectively. Find the width of the ring.

Solution:

Let $R$ and $r$ be the radii of the outer and inner circles respectively. Let $d$ be the width of the ring  $d = (R – r)$ $\therefore$ the area of the outer circle $= \pi {R^2} = 1886.5$ square cm
or  $R = 24.51$ cm
$\therefore$ the area of the inner circle $= \pi {r^2} = 1386$ square cm
$\therefore$ $\log \pi + 2\log r = \log 1386$
$2\log r = \log 1386 – \log (3.143)$
$= 3.1418 – 0.4973 = 2.6454$
$\log r = 1.3223$
$r = anti\log (1.3223) = 21$

Hence, the width of the ring $= R – r = 24.51 – 21 = 3.51$cm