Question Number 45366 by behi83417@gmail.com last updated on 12/Oct/18
Commented by behi83417@gmail.com last updated on 12/Oct/18
$${shaded}\:{area}=? \\ $$
Commented by ajfour last updated on 12/Oct/18
Answered by ajfour last updated on 12/Oct/18
$${x}=\:\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\theta\:=\:\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{180}°−\mathrm{20}°\right)×\frac{\pi}{\mathrm{180}°} \\ $$$$\:\:=\:\frac{\mathrm{9}}{\mathrm{2}}×\frac{\mathrm{8}\pi}{\mathrm{9}}\:=\:\mathrm{4}\boldsymbol{\pi}\: \\ $$$$\left({since}\:\angle{ACB}\:=\:\mathrm{150}°−\mathrm{130}°=\mathrm{20}°\:\right). \\ $$
Commented by behi83417@gmail.com last updated on 12/Oct/18
$${thanks}\:{a}\:{lot}\:{sir}\:{Ajfour}. \\ $$
Answered by MrW3 last updated on 12/Oct/18
![=For any three circles= Radius of circle: black=r_1 green=r_2 red=r_2 (r_1 −r_2 )^2 =(r_1 −r_3 )^2 +(r_2 +r_3 )^2 −2(r_1 −r_3 )(r_2 +r_3 ) cos θ cos θ=(((r_1 −r_3 )^2 +(r_2 +r_3 )^2 −(r_1 −r_2 )^2 )/(2(r_1 −r_3 )(r_2 +r_3 ))) cos θ=((r_3 ^2 −r_1 r_3 +r_2 r_3 +r_1 r_2 )/((r_1 −r_3 )(r_2 +r_3 ))) cos θ=((r_2 (r_1 +r_3 )−r_3 (r_1 −r_3 ))/((r_1 −r_3 )(r_2 +r_3 ))) θ=cos^(−1) {((r_2 (r_1 +r_3 )−r_3 (r_1 −r_3 ))/((r_1 −r_3 )(r_2 +r_3 )))} A_(shade) =((r_3 ^2 (π−θ))/2)=(r_3 ^2 /2)[π−cos^(−1) {((r_2 (r_1 +r_3 )−r_3 (r_1 −r_3 ))/((r_1 −r_3 )(r_2 +r_3 )))}]](Q45374.png)
$$={For}\:{any}\:{three}\:{circles}= \\ $$$${Radius}\:{of}\:{circle}:\: \\ $$$${black}={r}_{\mathrm{1}} \\ $$$${green}={r}_{\mathrm{2}} \\ $$$${red}={r}_{\mathrm{2}} \\ $$$$\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)\:\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\theta=\frac{\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)^{\mathrm{2}} −\left({r}_{\mathrm{1}} −{r}_{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}_{\mathrm{3}} ^{\mathrm{2}} −{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} {r}_{\mathrm{3}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} }{\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)−{r}_{\mathrm{3}} \left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)}{\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)−{r}_{\mathrm{3}} \left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)}{\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)}\right\} \\ $$$${A}_{{shade}} =\frac{{r}_{\mathrm{3}} ^{\mathrm{2}} \left(\pi−\theta\right)}{\mathrm{2}}=\frac{{r}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{2}}\left[\pi−\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{r}_{\mathrm{2}} \left({r}_{\mathrm{1}} +{r}_{\mathrm{3}} \right)−{r}_{\mathrm{3}} \left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)}{\left({r}_{\mathrm{1}} −{r}_{\mathrm{3}} \right)\left({r}_{\mathrm{2}} +{r}_{\mathrm{3}} \right)}\right\}\right] \\ $$
Commented by behi83417@gmail.com last updated on 12/Oct/18
$${thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$$${how}\:{can}\:{we}\:{find}\:\:{r}_{\mathrm{1}} \:{and}\:{r}_{\mathrm{2}} ? \\ $$
Commented by MrW3 last updated on 12/Oct/18
Commented by MrW3 last updated on 12/Oct/18
$${this}\:{shows}\:{the}\:{question}\:{I}\:{solved}.\:{it} \\ $$$${is}\:{not}\:{directly}\:{your}\:{question}. \\ $$