Tag: area between two concentric circles

Area of equilateral triangle $= s$ sq.cm
$\Rightarrow  \dfrac{\sqrt3}{4} a^2 = s$, [where $a$, the side of equilateral triangle]
$\Rightarrow a= \sqrt{\dfrac{4s}{\sqrt3}}$
Now perimeter of equilateral triangle $ 3\times a =3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$ cm 
Circumference of circle $=$ perimeter of equilateral triangle
$\Rightarrow 2\pi r= 3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$, [where $r$ the radius of circle]
Solve the above expression for $r$, we get 
$r= \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}}$
Area of circle $=\pi r^2 = \pi \times \left ( \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}} \right )^2$
After simplification, we get
Area of circle $=\dfrac{3s\sqrt3}{\pi}$ sq.cm

Let the number of revolution of the wheel is n.
Then,
n $\times$ circumference of wheel = Distance travelled by bicycle
$n \times  2\Pi  \times \frac {1}{2}=1$ kilometer 
$n \times  \Pi $=1000 meter
$n=\frac {1000}{\Pi }$

A dog is chained on a $6$ ft leash to the corner of a rectangular building.
Since the building is rectangular, the dog is left with an angle of $360 - 90 = 270^o$ for it to roam around.
Also, the length of the leash will act as the radius of this sector.
$\therefore$ Area of the sector $= \cfrac{270}{360} \times \pi \times 6^2$
$= \cfrac{3}{4} \times \pi \times 36$
$= 84.82 \ \ ft^2$