Find the ratio of two different sector of a circle with radius 20 cm a

1.	Find the ratio of two different sector of a circle with radius 20 cm and angles respectively 15 and 90.
Answer» $\large\underline{\sf{Solution-}}$ Dimensions of first SECTOR :- $\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{radius,r_1 = 20 \: cm} \\ &\sf{angle,\theta_1 \: = 90 \degree} \end{cases}\end{gathered}\end{gathered}$ We know, AREA of sector is $\boxed{ \sf \: Area_{(sector)} = \pi \: {<klux>R</klux>}^{2} \dfrac{\theta}{360\degree}}$ where, r is the radius of sector θ is central angle or sector angle. So, on substituting the VALUES of r and θ, we get $\rm :\longmapsto\:Area_{(sector1)} = \pi \: {(20)}^{2} \dfrac{90\degree}{360\degree}$ $\bf\implies \:Area_{(sector1)} = 100 \: \pi \: {cm}^{2} - - - (1)$ Now, Dimensions of second sector :- $\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{radius,r_2 = 20 \: cm} \\ &\sf{angle,\theta_2 \: = 15 \degree} \end{cases}\end{gathered}\end{gathered}$ So, again on substituting the values of r and θ, we get $\rm :\longmapsto\:Area_{(sector2)} = \pi \: {(20)}^{2} \dfrac{15\degree}{360\degree}$ $\bf\implies \:Area_{(sector2)} = \dfrac{50\pi}{3}\:{cm}^{2} - - - (2)$ Therefore, $\rm :\longmapsto\:\dfrac{Area_{(sector1)}}{Area_{(sector2)}} = \dfrac{100\pi}{\dfrac{50\pi}{3} }$ $\rm :\longmapsto\:\dfrac{Area_{(sector1)}}{Area_{(sector2)}} = 6$ $\bf\implies \:Area_{(sector1)}: Area_{(sector2)} = 6 : 1$ ─━─━─━─━─━─━─━─━─━─━─━─━─ Additional INFORMATION :- $\boxed{ \sf \: Length \: of \: arc_{(sector)} = 2\pi \: r \: \dfrac{\theta}{360\degree}}$ $\boxed{ \sf \: Perimeter_{(sector)} = 2r + l}$ $\boxed{ \sf \: Area_{(major \: sector)} = \pi \: {r}^{2} \dfrac{(360\degree - \theta)}{360\degree}}$ $\boxed{ \sf \: Area_{(minor \: segment)} = \bigg[\pi \: {r}^{2} \dfrac{\theta}{360\degree} - \dfrac{1}{2} {r}^{2}sin\theta\bigg]}$

Find the ratio of two different sector of a circle with radius 20 cm and angles respectively 15 and 90.

Answer»

$\large\underline{\sf{Solution-}}$

Dimensions of first SECTOR :-

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{radius,r_1 = 20 \: cm} \\ &\sf{angle,\theta_1 \: = 90 \degree} \end{cases}\end{gathered}\end{gathered}$

We know,

AREA of sector is

$\boxed{ \sf \: Area_{(sector)} = \pi \: {<klux>R</klux>}^{2} \dfrac{\theta}{360\degree}}$

where,

r is the radius of sector

θ is central angle or sector angle.

So, on substituting the VALUES of r and θ, we get

$\rm :\longmapsto\:Area_{(sector1)} = \pi \: {(20)}^{2} \dfrac{90\degree}{360\degree}$

$\bf\implies \:Area_{(sector1)} = 100 \: \pi \: {cm}^{2} - - - (1)$

Now,

Dimensions of second sector :-

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{radius,r_2 = 20 \: cm} \\ &\sf{angle,\theta_2 \: = 15 \degree} \end{cases}\end{gathered}\end{gathered}$

So, again on substituting the values of r and θ, we get

$\rm :\longmapsto\:Area_{(sector2)} = \pi \: {(20)}^{2} \dfrac{15\degree}{360\degree}$

$\bf\implies \:Area_{(sector2)} = \dfrac{50\pi}{3}\:{cm}^{2} - - - (2)$

Therefore,

$\rm :\longmapsto\:\dfrac{Area_{(sector1)}}{Area_{(sector2)}} = \dfrac{100\pi}{\dfrac{50\pi}{3} }$

$\rm :\longmapsto\:\dfrac{Area_{(sector1)}}{Area_{(sector2)}} = 6$

$\bf\implies \:Area_{(sector1)}: Area_{(sector2)} = 6 : 1$

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Additional INFORMATION :-

$\boxed{ \sf \: Length \: of \: arc_{(sector)} = 2\pi \: r \: \dfrac{\theta}{360\degree}}$

$\boxed{ \sf \: Perimeter_{(sector)} = 2r + l}$

$\boxed{ \sf \: Area_{(major \: sector)} = \pi \: {r}^{2} \dfrac{(360\degree - \theta)}{360\degree}}$

$\boxed{ \sf \: Area_{(minor \: segment)} = \bigg[\pi \: {r}^{2} \dfrac{\theta}{360\degree} - \dfrac{1}{2} {r}^{2}sin\theta\bigg]}$