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r = 12 cm, {tex}\theta{/tex} = 120o
{tex}\therefore{/tex} Area of the corresponding sector of the circle = {tex}\frac\theta{360^\circ}\mathrm{πr}^2\;=\;\frac{\displaystyle120}{\displaystyle360^\circ}\times3.14\;\times12\times12\;{/tex}= 150.72 cm2
Area of {tex}\triangle{/tex}AOB
Draw OM {tex}\perp{/tex} AB
In right triangle OMA and OMB,
OA = OB ....... Radii of the same circle
OM = OM ........ Common side
{tex}\therefore{/tex} {tex}\triangle{/tex}OMA {tex}\cong{/tex} {tex}\triangle{/tex}OMB ........RHS congruence criterion
{tex}\therefore{/tex} AM = BM ....... CPCT
{tex}\Rightarrow{/tex} AM = BM = {tex}\frac 12{/tex}AB
and {tex}\angle{/tex}AOM = {tex}\angle{/tex}BOM [CPCT]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}AOM = {tex}\angle{/tex}BOM = {tex}\frac 12{/tex}{tex}\angle{/tex}AOB
= {tex}\frac 12{/tex} {tex}\times{/tex} 120o = 60o
{tex}\therefore{/tex} In right triangle OMA,
cos60o = {tex}\frac {OM}{OA}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac 12{/tex}= {tex}\frac {OM}{12}{/tex}
{tex}\Rightarrow{/tex} OM = 6 cm
sin60o = {tex}\frac {AM}{OA}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{\sqrt3}2{/tex}= {tex}\frac {AM}{12}{/tex}
{tex}\Rightarrow{/tex} AM = 6{tex}\sqrt3{/tex} cm
{tex}\Rightarrow{/tex} 2AM = 12{tex}\sqrt3{/tex} cm
{tex}\Rightarrow{/tex} AB =12{tex}\sqrt3{/tex} cm
{tex}\therefore{/tex} Area of {tex}\triangle{/tex}AOB ={tex}\frac 12{/tex} {tex}\times{/tex} AB {tex}\times{/tex} OM
= {tex}\frac 12{/tex} {tex}\times{/tex} 12{tex}\sqrt3{/tex} {tex}\times{/tex} 6 = 36{tex}\sqrt3{/tex} cm2
= 36 {tex}\times{/tex} 1.73 cm2 = 62.28 cm2
So, Area of the corresponding segment of the circle = Area of the correspoding sector of circle - Area of {tex}\triangle{/tex}AOB
= 150.72 - 62.28 = 88.44 cm2
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M = m / n
M = molar mass
m = mass of a substance (in grams)
n = number of moles of a substance
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