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Question Number 57184 by Tawa1 last updated on 31/Mar/19

Commented by Tawa1 last updated on 31/Mar/19

Find the area of the shaded portion

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{portion} \\ $$

Answered by mr W last updated on 31/Mar/19

R=radius of big circle=5 r=radius of small circle (√2)r+r+R=(√2)R ⇒r=((((√2)−1)R)/((√2)+1))=(3−2(√2))R blue area =R^2 −((πR^2 )/4)−πr^2 =R^2 −((πR^2 )/4)−π(17−12(√2))R^2 =((4−(48(√2)−67)π)/4)R^2 =0.307R^2 =7.677 cm^2

$${R}={radius}\:{of}\:{big}\:{circle}=\mathrm{5} \\ $$$${r}={radius}\:{of}\:{small}\:{circle} \\ $$$$\sqrt{\mathrm{2}}{r}+{r}+{R}=\sqrt{\mathrm{2}}{R} \\ $$$$\Rightarrow{r}=\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){R}}{\sqrt{\mathrm{2}}+\mathrm{1}}=\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right){R} \\ $$$${blue}\:{area}\:={R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} \\ $$$$={R}^{\mathrm{2}} −\frac{\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\pi\left(\mathrm{17}−\mathrm{12}\sqrt{\mathrm{2}}\right){R}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}−\left(\mathrm{48}\sqrt{\mathrm{2}}−\mathrm{67}\right)\pi}{\mathrm{4}}{R}^{\mathrm{2}} \\ $$$$=\mathrm{0}.\mathrm{307}{R}^{\mathrm{2}} \\ $$$$=\mathrm{7}.\mathrm{677}\:{cm}^{\mathrm{2}} \\ $$

Commented by mr W last updated on 31/Mar/19

Commented by Tawa1 last updated on 31/Mar/19

God bless you sir. But sir, i have a request. I want to understand your steps sir. please help me explain some steps.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$$$\mathrm{But}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{have}\:\mathrm{a}\:\mathrm{request}.\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{steps}\:\mathrm{sir}. \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{some}\:\mathrm{steps}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Mar/19

a=2R area of shaded portion =((a^2 −πR^2 )/4)−πr^2 =((a^2 −π((a/2))^2 )/4)−πr^2 now diagonal of square=a(√2) a(√2) =2R+4r+2l (l+r)^2 =r^2 +r^2 l+r=r(√2) l=r((√2) −1) a(√2) =2R+4r+2l a(√2) =a+4r+2r((√2) −1) a(√2) −a=2r+2(√2) r a((√2) −1)=r(2+2(√2) ) r=((a((√2) −1))/(2((√2) +1))) so shaded area =((a^2 −π((a/2))^2 )/4)−πr^2 =((a^2 −π((a/2))^2 )/4)−π×{((a((√2) −1))/(2((√2) +1))}^2 now put a=10 pls check mistake if any

$${a}=\mathrm{2}{R} \\ $$$${area}\:{of}\:{shaded}\:{portion} \\ $$$$=\frac{{a}^{\mathrm{2}} −\pi{R}^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} −\pi\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} \\ $$$${now}\:{diagonal}\:{of}\:{square}={a}\sqrt{\mathrm{2}}\: \\ $$$${a}\sqrt{\mathrm{2}}\:=\mathrm{2}{R}+\mathrm{4}{r}+\mathrm{2}{l} \\ $$$$\left({l}+{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$${l}+{r}={r}\sqrt{\mathrm{2}}\: \\ $$$${l}={r}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$$${a}\sqrt{\mathrm{2}}\:=\mathrm{2}{R}+\mathrm{4}{r}+\mathrm{2}{l} \\ $$$${a}\sqrt{\mathrm{2}}\:={a}+\mathrm{4}{r}+\mathrm{2}{r}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right) \\ $$$${a}\sqrt{\mathrm{2}}\:−{a}=\mathrm{2}{r}+\mathrm{2}\sqrt{\mathrm{2}}\:{r} \\ $$$${a}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)={r}\left(\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\:\right) \\ $$$${r}=\frac{{a}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)} \\ $$$${so}\:{shaded}\:{area} \\ $$$$=\frac{{a}^{\mathrm{2}} −\pi\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}−\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} −\pi\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{4}}−\pi×\left\{\frac{{a}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)}{\mathrm{2}\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right.}\right\}^{\mathrm{2}} \\ $$$${now}\:{put}\:{a}=\mathrm{10} \\ $$$${pls}\:{check}\:{mistake}\:{if}\:{any} \\ $$

Commented by Tawa1 last updated on 31/Mar/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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