Area of Circular Rings (Annuli) - Class IX

Area of outer circle $=\pi{(19)}^2$
                                 $=\cfrac{22}{7}\times 361$
Area of inner circle $=\pi{(16)}^2$
                                 $=\cfrac{22}{7}\times 256$
$\therefore$ area of the ring $=\dfrac{22}{7}\times 361-\dfrac{22}{7}\times 256$
                              $=\cfrac{22}{7}(361-256)$
                              $=\cfrac{22}{7}(105)$
                              $=330\ cm^2$

Let radius of the inner circle be $r$ and radius of the outer circle be $R=21cm$

Given: $r _1=5+r _2$
$\displaystyle \frac {2\pi r _1}{2\pi r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow \frac {r _1}{r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow \frac {5+r _2}{r _2}=\frac {23}{22}$

$\displaystyle \Rightarrow 110+22r _2=23r _2$

$\displaystyle \Rightarrow r _2=110\ m$
$\therefore $ Radius of inner circle$=110\ m$

 Area of the circular ring = $\frac { 22 }{ 7 } \times \left( { R }^{ 2 } - { r }^{ 2 } \right)$ = $ \frac { 22 }{ 7 } \times \left( { 14 }^{ 2 } - { 10.5 }^{ 2 } \right)$ = $269.5 \ { cm }^{ 2 }\approx 269\ { cm }^{ 2 } $

$Area=\pi (100^{2} - 8^{2}) = \pi 36 = 113.143\ cm^{2}$.

The area of a ring shaped region$=\pi(20)^2-\pi(15)^2$
                                                    $=(400-225)\times \dfrac{22}{7}$
                                                    $=(175)\times \dfrac{22}{7}$
                                                   $ =550$ sq. cm

Given $\displaystyle 2\pi r _{2}-2\pi r _{1}=132$
$\displaystyle \Rightarrow r _{2}-r _{1}=\frac{132}{2\pi }=\frac{132\times 7}{44}=21\ m$

Inner circumference $= \displaystyle 2\pi r=2\times \frac{22}{7}\times r=\frac{44r}{7}$
Given $\displaystyle \frac{44r}{7}=440\Rightarrow r=\frac{440\times 7}{44}=70\ m$
$\displaystyle \therefore $ Radius of outer circle $= 70\ m + 14\ m = 84\ m$ 

$R = 15\ cm, r = 13\ cm$
Area of the circulating ring 
$\displaystyle = \pi \left(R^2-r^2\right)$

Circumference of inner side = $220 m$

Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
Given the area of outer circle=616$\displaystyle cm^{2}$
$\displaystyle \Rightarrow  \pi r _{2}^{2}=616 cm^{2}$
$\displaystyle \Rightarrow r _{1}^{2}=\frac{616\times 7}{22}=196$
$\displaystyle \therefore r _{1}=14 cm $
and the area of the inner circle $\displaystyle =154 cm^{2}$
$\displaystyle \Rightarrow \pi r _{2}^{2}= 154$
$\displaystyle \Rightarrow r _{2}^{2}=\frac{154\times 7}{22}=49$
$\displaystyle \therefore r _{2}=7 cm.$
$\displaystyle \therefore $ The required answer $\displaystyle =r _{1}-r _{2}=14-7=7 cm.$

Area of a circular ring $=\pi(R^2-r^2)$
$=3.14(12^2-10^2)$
$= 3.14 \times 44$
$= 138.16$ $cm^2$

Given : Radius ($r _1$) $=21m$

Radius of bigger circle(with the path) = $35 + 7 = 42\ m.$
Thus area of the path $=$ Area of bigger circle $-$ Area of smaller circle
$\therefore$ Required area $= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$