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Question-103347

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Question Number 103347 by I want to learn more last updated on 14/Jul/20
Commented by I want to learn more last updated on 14/Jul/20
Commented by Tawa11 last updated on 15/Sep/21
nice
$$\mathrm{nice} \\ $$
Answered by 1549442205 last updated on 14/Jul/20
R_1 =12cm,R_2 =8cm,r=2 cm The length of the arc QR is p_1 =((2πR_1 )/(360))×80=((2π×12×8)/(36))=((16π)/3)(cm) The length of the arc PS is p_2 =((2πR_2 )/(360))×80=((2π×8×8)/(36))=((32π)/9)(cm) The length of two semicircles is p_3 =2πr=4π(cm) The perimeter of shape is P=p_1 +p_2 +p_3 =((16𝛑)/3)+((32𝛑)/9)+4𝛑=((116𝛑)/9)(cm) b)The area of the sector OQR is S_1 =πR_1 ^2 ×((80)/(360))=((144×8π)/(36))=72π(cm^2 ) The area of the sector OPS is S_2 =πR_2 ^2 ×((80)/(360))=((64×8π)/(36))=((128π)/9) (cm^2 ) The area of two semicircles is S_3 =πr^2 =4π(cm^2 ) The are of shape is S=(S_1 −S_2 )+S_3 =72𝛑−((128𝛑)/9)−4𝛑=((484𝛑)/9)(cm^2 )
$$\mathrm{R}_{\mathrm{1}} =\mathrm{12cm},\mathrm{R}_{\mathrm{2}} =\mathrm{8cm},\mathrm{r}=\mathrm{2}\:\mathrm{cm} \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{QR}\:\mathrm{is}\:\:\mathrm{p}_{\mathrm{1}} =\frac{\mathrm{2}\pi\mathrm{R}_{\mathrm{1}} }{\mathrm{360}}×\mathrm{80}=\frac{\mathrm{2}\pi×\mathrm{12}×\mathrm{8}}{\mathrm{36}}=\frac{\mathrm{16}\pi}{\mathrm{3}}\left(\mathrm{cm}\right) \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{PS}\:\mathrm{is}\:\:\mathrm{p}_{\mathrm{2}} =\frac{\mathrm{2}\pi\mathrm{R}_{\mathrm{2}} }{\mathrm{360}}×\mathrm{80}=\frac{\mathrm{2}\pi×\mathrm{8}×\mathrm{8}}{\mathrm{36}}=\frac{\mathrm{32}\pi}{\mathrm{9}}\left(\mathrm{cm}\right) \\ $$$$\mathrm{The}\:\mathrm{length}\:\mathrm{of}\:\mathrm{two}\:\mathrm{semicircles}\:\mathrm{is}\:\mathrm{p}_{\mathrm{3}} =\mathrm{2}\pi\mathrm{r}=\mathrm{4}\pi\left(\mathrm{cm}\right) \\ $$$$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{shape}\:\mathrm{is} \\ $$$$\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{p}}_{\mathrm{1}} +\boldsymbol{\mathrm{p}}_{\mathrm{2}} +\boldsymbol{\mathrm{p}}_{\mathrm{3}} =\frac{\mathrm{16}\boldsymbol{\pi}}{\mathrm{3}}+\frac{\mathrm{32}\boldsymbol{\pi}}{\mathrm{9}}+\mathrm{4}\boldsymbol{\pi}=\frac{\mathrm{116}\boldsymbol{\pi}}{\mathrm{9}}\left(\boldsymbol{\mathrm{cm}}\right) \\ $$$$\left.\mathrm{b}\right)\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sector}\:\mathrm{OQR}\:\mathrm{is} \\ $$$$\mathrm{S}_{\mathrm{1}} =\pi\mathrm{R}_{\mathrm{1}} ^{\mathrm{2}} ×\frac{\mathrm{80}}{\mathrm{360}}=\frac{\mathrm{144}×\mathrm{8}\pi}{\mathrm{36}}=\mathrm{72}\pi\left(\mathrm{cm}^{\mathrm{2}} \right) \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sector}\:\mathrm{OPS}\:\mathrm{is} \\ $$$$\mathrm{S}_{\mathrm{2}} =\pi\mathrm{R}_{\mathrm{2}} ^{\mathrm{2}} ×\frac{\mathrm{80}}{\mathrm{360}}=\frac{\mathrm{64}×\mathrm{8}\pi}{\mathrm{36}}=\frac{\mathrm{128}\pi}{\mathrm{9}}\:\left(\mathrm{cm}^{\mathrm{2}} \right) \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{two}\:\mathrm{semicircles}\:\mathrm{is} \\ $$$$\mathrm{S}_{\mathrm{3}} =\pi\mathrm{r}^{\mathrm{2}} =\mathrm{4}\pi\left(\mathrm{cm}^{\mathrm{2}} \right) \\ $$$$\mathrm{The}\:\mathrm{are}\:\mathrm{of}\:\mathrm{shape}\:\mathrm{is}\: \\ $$$$\boldsymbol{\mathrm{S}}=\left(\boldsymbol{\mathrm{S}}_{\mathrm{1}} −\boldsymbol{\mathrm{S}}_{\mathrm{2}} \right)+\boldsymbol{\mathrm{S}}_{\mathrm{3}} =\mathrm{72}\boldsymbol{\pi}−\frac{\mathrm{128}\boldsymbol{\pi}}{\mathrm{9}}−\mathrm{4}\boldsymbol{\pi}=\frac{\mathrm{484}\boldsymbol{\pi}}{\mathrm{9}}\left(\boldsymbol{\mathrm{cm}}^{\mathrm{2}} \right) \\ $$$$ \\ $$
Commented by I want to learn more last updated on 14/Jul/20
Thanks sir, i appreciate
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate} \\ $$

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