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Question Number 135827 by otchereabdullai@gmail.com last updated on 16/Mar/21

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{leaving}\mathrm{the}$$$$\mathrm{answer}\mathrm{in}\pi .$$$$\mathrm{i})\mathrm{if}\mathrm{it}\mathrm{area}\mathrm{is}\mathrm{doubled}\mathrm{it}\mathrm{circumference}$$$$$$$$\mathrm{ii})\mathrm{if}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{it}\mathrm{cermi}-\mathrm{circle}\mathrm{is}$$$$\mathrm{numerically}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{arc}\mathrm{length}\mathrm{of}$$$$\mathrm{a}\mathrm{quater}\mathrm{circle}$$$$$$$$\mathrm{iii})\mathrm{if}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{a}\mathrm{quater}\mathrm{circle}\mathrm{is}$$$$\mathrm{numerically}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{lenght}\mathrm{of}\mathrm{an}$$$$\mathrm{arc}\mathrm{subtendig}\mathrm{to}\mathrm{an}\mathrm{angle}\mathrm{of}\frac{2\pi }{3}\mathrm{radian}$$$$\mathrm{at}\mathrm{the}\mathrm{centre}.$$$$\mathrm{please}\mathrm{i}\mathrm{need}\mathrm{help}$$

Answered by mr W last updated on 16/Mar/21

$$i)$$$$\pi r^2=2\times 2\pi r$$$$\Rightarrow r=4$$$$areaofcircle=\pi r^2=16\pi$$$$ii)$$$$\frac{\pi r^2}{2}=\frac{2\pi r}{4}$$$$\Rightarrow r=1$$$$areaofcircle=\pi r^2=\pi$$$$iii)$$$$\frac{\pi r^2}{4}=\frac{2\pi }{3}\times r$$$$\Rightarrow r=\frac{8}{3}$$$$areaofcircle=\pi r^2=\frac{64}{9}\pi$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/21

$$\mathrm{ProfW}\mathrm{your}\mathrm{reward}\mathrm{is}\mathrm{in}\mathrm{heaven}\mathrm{thank}$$$$\mathrm{you}\mathrm{for}\mathrm{being}\mathrm{so}\mathrm{kind}\mathrm{to}\mathrm{me}!$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/21

$$\mathrm{prof}\mathrm{please}\mathrm{check}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{first}$$$$\mathrm{one}\left(\mathrm{i}\right)\mathrm{for}\mathrm{me}.\mathrm{the}\mathrm{question}\mathrm{says}\mathrm{if}\mathrm{it}$$$$\mathrm{area}\mathrm{is}\mathrm{doubled}\mathrm{it}\mathrm{circumference}\mathrm{so}\mathrm{i}\mathrm{thing}\mathrm{it}\mathrm{should}\mathrm{be}$$$$\pi {\mathrm{r}}^2=2\left(2\pi \mathrm{r}\right)\mathrm{so}\mathrm{that}\mathrm{r}=4\mathrm{and}\mathrm{final}$$$$\mathrm{answer}\mathrm{to}\mathrm{be}16\pi$$$$\mathrm{i}\mathrm{did}\mathrm{not}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{aproach}\mathrm{it}$$$$\mathrm{but}\mathrm{i}\mathrm{got}\mathrm{the}\mathrm{idea}\mathrm{through}\mathrm{your}\mathrm{solution}$$$$\mathrm{so}\mathrm{my}\mathrm{suggestion}$$

Commented by mr W last updated on 16/Mar/21

$$youareright.ioverlooked.$$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/21

$$\mathrm{thank}\mathrm{you}\mathrm{prof}\mathrm{W}$$

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