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Question Number 135827 by otchereabdullai@gmail.com last updated on 16/Mar/21
Find the area of a circle leaving the answer in π. i) if it area is doubled it circumference ii) if the area of it cermi-circle is numerically equal to the arc length of a quater circle iii) if the area of a quater circle is numerically equal to the lenght of an arc subtendig to an angle of ((2π)/3) radian at the centre. please i need help
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{leaving}\:\mathrm{the}\: \\ $$$$\mathrm{answer}\:\mathrm{in}\:\pi. \\ $$$$\left.\mathrm{i}\right)\:\mathrm{if}\:\mathrm{it}\:\mathrm{area}\:\mathrm{is}\:\mathrm{doubled}\:\mathrm{it}\:\mathrm{circumference} \\ $$$$ \\ $$$$\left.\mathrm{ii}\right)\:\mathrm{if}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{it}\:\mathrm{cermi}-\mathrm{circle}\:\mathrm{is}\: \\ $$$$\mathrm{numerically}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{length}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{quater}\:\mathrm{circle} \\ $$$$ \\ $$$$\left.\mathrm{iii}\right)\:\mathrm{if}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{quater}\:\mathrm{circle}\:\mathrm{is}\: \\ $$$$\mathrm{numerically}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{arc}\:\mathrm{subtendig}\:\mathrm{to}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:\mathrm{radian} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{centre}. \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help} \\ $$
Answered by mr W last updated on 16/Mar/21
i) πr^2 =2×2πr ⇒r=4 area of circle=πr^2 =16π ii) ((πr^2 )/2)=((2πr)/4) ⇒r=1 area of circle=πr^2 =π iii) ((πr^2 )/4)=((2π)/3)×r ⇒r=(8/3) area of circle=πr^2 =((64)/9)π
$$\left.{i}\right) \\ $$$$\pi{r}^{\mathrm{2}} =\mathrm{2}×\mathrm{2}\pi{r} \\ $$$$\Rightarrow{r}=\mathrm{4} \\ $$$${area}\:{of}\:{circle}=\pi{r}^{\mathrm{2}} =\mathrm{16}\pi \\ $$$$\left.{ii}\right) \\ $$$$\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}\pi{r}}{\mathrm{4}} \\ $$$$\Rightarrow{r}=\mathrm{1} \\ $$$${area}\:{of}\:{circle}=\pi{r}^{\mathrm{2}} =\pi \\ $$$$\left.{iii}\right) \\ $$$$\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{2}\pi}{\mathrm{3}}×{r} \\ $$$$\Rightarrow{r}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$${area}\:{of}\:{circle}=\pi{r}^{\mathrm{2}} =\frac{\mathrm{64}}{\mathrm{9}}\pi \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Mar/21
ProfW your reward is in heaven thank you for being so kind to me!
$$\mathrm{ProfW}\:\mathrm{your}\:\mathrm{reward}\:\mathrm{is}\:\mathrm{in}\:\mathrm{heaven}\:\mathrm{thank} \\ $$$$\mathrm{you}\:\mathrm{for}\:\mathrm{being}\:\mathrm{so}\:\mathrm{kind}\:\mathrm{to}\:\mathrm{me}! \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Mar/21
prof please check solution to the first one (i) for me. the question says if it area is doubled it circumference so i thing it should be πr^2 =2(2πr) so that r=4 and final answer to be 16π i did not know how to aproach it but i got the idea through your solution so my suggestion
$$\mathrm{prof}\:\mathrm{please}\:\mathrm{check}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{one}\:\left(\mathrm{i}\right)\:\:\mathrm{for}\:\mathrm{me}.\:\mathrm{the}\:\mathrm{question}\:\mathrm{says}\:\mathrm{if}\:\mathrm{it}\: \\ $$$$\mathrm{area}\:\mathrm{is}\:\mathrm{doubled}\:\mathrm{it}\:\mathrm{circumference}\:\mathrm{so}\:\mathrm{i}\:\mathrm{thing}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$$\pi\mathrm{r}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{2}\pi\mathrm{r}\right)\:\mathrm{so}\:\mathrm{that}\:\mathrm{r}=\mathrm{4}\:\mathrm{and}\:\mathrm{final}\: \\ $$$$\mathrm{answer}\:\mathrm{to}\:\mathrm{be}\:\mathrm{16}\pi\: \\ $$$$\mathrm{i}\:\mathrm{did}\:\mathrm{not}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{aproach}\:\mathrm{it}\: \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{got}\:\mathrm{the}\:\mathrm{idea}\:\mathrm{through}\:\mathrm{your}\:\mathrm{solution} \\ $$$$\mathrm{so}\:\mathrm{my}\:\mathrm{suggestion}\: \\ $$
Commented by mr W last updated on 16/Mar/21
you are right. i overlooked.
$${you}\:{are}\:{right}.\:{i}\:{overlooked}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 16/Mar/21
thank you prof W
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{prof}\:\mathrm{W} \\ $$

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