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Question-177013




Question Number 177013 by mr W last updated on 29/Sep/22
Commented by mr W last updated on 29/Sep/22
find the area of the sector.
$${find}\:{the}\:{area}\:{of}\:{the}\:{sector}. \\ $$
Answered by Rasheed.Sindhi last updated on 29/Sep/22
The diameter=(√(4^2 +(3+5)^2 )) =4(√5)  Rsdius=2(√5)  Area of circle=π(2(√5) )^2 =20π  Area of sector=((20π)/4)=5π
$$\mathcal{T}{he}\:{diameter}=\sqrt{\mathrm{4}^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} }\:=\mathrm{4}\sqrt{\mathrm{5}} \\ $$$${Rsdius}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${Area}\:{of}\:{circle}=\pi\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} =\mathrm{20}\pi \\ $$$${Area}\:{of}\:{sector}=\frac{\mathrm{20}\pi}{\mathrm{4}}=\mathrm{5}\pi \\ $$
Commented by mr W last updated on 29/Sep/22
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Commented by Rasheed.Sindhi last updated on 29/Sep/22
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Commented by mr W last updated on 29/Sep/22
Commented by Rasheed.Sindhi last updated on 29/Sep/22
Thanks for diagram sir!  🙏🌷🌹🙏
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{for}\:\mathrm{diagram}\:\boldsymbol{\mathrm{sir}}! \\ $$🙏🌷🌹🙏
Commented by mr W last updated on 30/Sep/22
thanks for smart solution sir!
$${thanks}\:{for}\:{smart}\:{solution}\:{sir}! \\ $$
Commented by Tawa11 last updated on 02/Oct/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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