Length of an arc - class-IX

$\displaystyle \frac{36}{360}\times \frac{22}{7}r^{2}=15.4\Rightarrow r^{2} =\frac{15.4\times 5\times 7}{11}=49$
$\displaystyle \Rightarrow r=7$
$\displaystyle C=2\pi r=2\times \frac{22}{7}\times 7=44cm$
$\displaystyle =0.44 m$

$\displaystyle \theta =56^{\circ},r=21cm$
Length. of $\displaystyle AB=\dfrac{56^{\circ}}{360^{\circ}}\times 2\times \dfrac{22}{7}\times 21$
$\displaystyle =\dfrac{616}{30}=20.53cm$

Setting a proportion 
AB : OB : : 126 : 360
$\displaystyle \frac{\overline{AB}}{2\pi r}=\frac{126}{360}$
$\displaystyle \overline{AB}=\left ( \frac{126}{360} \right )\times 2\pi r$
$\displaystyle \overline{AB}=\left ( \frac{126}{360} \right )\times 2\pi \times 8$
$\displaystyle \overline{AB}=5.6\pi $

Perimeter of the Sector = 2(Radius of the circle) + Arc Length 

Arc length = (theta / 360) * 2 * pi * r. Here, (60 / 360) * 2 * (22 / 7) * 14 = (1 / 6) * 2 * 22 * 2 = 88 / 6 = 44 / 3 cm. The statement is true.

Arc length = (60 / 360) * 2 * (22 / 7) * 21 = (1 / 6) * 2 * 22 * 3 = 132 / 6 = 22 cm. The statement is true.

The formula provided, (theta / 360) * pi * r^2, is the formula for the area of a sector, not the length of an arc. The arc length formula is (theta / 360) * 2 * pi * r.

Length of an arc of the sector $=r\times \theta$

Radius of the circle $ = \dfrac {\text{Diameter}}{2} = 5 $ cm 

Radius of the circle $ = \dfrac {Diameter}{2} = 5  cm $

Length of an arc subtending an angle $ \theta  = \dfrac { \theta  }{ 360 }

\times 2\pi R $ where R is the radius of the circle. 

The length of arc $l$ and area $A$ of a sector of angle $\theta$ in a circle of radius $r$ are given by,

$\text{Since we know that angle in radian }\theta =\dfrac{arc length}{radius}$
Here, $\theta=72^0=\dfrac{\pi}{180}\times72=\dfrac{2\pi}{5}$
and arc length=88 m

$l+2r=18$
$\displaystyle\frac{lr}{2}=20$
$lr=40$
$l=\displaystyle\frac{40}{r}$
$18=\displaystyle\frac{40}{r}+2r$
$r^2-9r+20=0$
$(r-4);;(r-5)=0$
$r=4$ or $r=5$
$l=8$ or $l=10;cm$

length of arc = $\dfrac { \theta  }{ { 360 }^{ 0 } } \times 2\pi r=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times 2\times \dfrac { 22 }{ 7 } \times 21=44cm$

We know that length of subtended arc $=$ $\theta r$
Here, $\theta = \cfrac{2\pi}{3}$
$r = 12$ cm
So, $ s = \cfrac{2\pi}{3}\times 12$
$s = 8\pi$ cm

Area of sector = $\dfrac{RL}{2}$
$230=\dfrac{11L}{2}$
$L = 41.81 cm$

Distance covered  $=\dfrac { \theta  }{ 360 } \times 2\pi r=\dfrac { 105 }{ 360 } \times 2\times \dfrac { 22 }{ 7 } \times 30=55m$

Length of an arc of a circle $=\dfrac { \theta  }{ 360 } \times 2\pi { r }=\dfrac { 144 }{ 360 } \times 2\pi \times \dfrac { 10 }{ 2 } =4\pi $ cm 

Given, radius of circular wire $= 7$ cm