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Question-138554




Question Number 138554 by cherokeesay last updated on 14/Apr/21
Answered by mr W last updated on 14/Apr/21
Commented by mr W last updated on 14/Apr/21
r=3  θ=20°  ΔBCD=((ΔACD)/2)=ΔCDE  shaded area=sector CED=r^2 θ                            =3^2 ×((20π)/(180))=π
$${r}=\mathrm{3} \\ $$$$\theta=\mathrm{20}° \\ $$$$\Delta{BCD}=\frac{\Delta{ACD}}{\mathrm{2}}=\Delta{CDE} \\ $$$${shaded}\:{area}={sector}\:{CED}={r}^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}^{\mathrm{2}} ×\frac{\mathrm{20}\pi}{\mathrm{180}}=\pi \\ $$
Commented by cherokeesay last updated on 14/Apr/21
but in the sector CED there is 2θ ???
$${but}\:{in}\:{the}\:{sector}\:{CED}\:{there}\:{is}\:\mathrm{2}\theta\:??? \\ $$$$ \\ $$
Commented by mr W last updated on 15/Apr/21
area of sector =((r^2 ×angle)/2)=((r^2 ×2θ)/2)=r^2 θ
$${area}\:{of}\:{sector}\:=\frac{{r}^{\mathrm{2}} ×{angle}}{\mathrm{2}}=\frac{{r}^{\mathrm{2}} ×\mathrm{2}\theta}{\mathrm{2}}={r}^{\mathrm{2}} \theta \\ $$
Commented by cherokeesay last updated on 15/Apr/21
thanks
$${thanks} \\ $$

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