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Question-86256




Question Number 86256 by TawaTawa1 last updated on 27/Mar/20
Answered by ajfour last updated on 27/Mar/20
πr^2 =πa^2 (3+2(√2))
$$\pi{r}^{\mathrm{2}} =\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$
Answered by ajfour last updated on 27/Mar/20
Commented by ajfour last updated on 27/Mar/20
area of △= (s^2 /2)= a^2   (given)  AB+BC = (s/( (√2)))+r = r(√2)  r=(s/(2−(√2))) = (s/2)(2+(√2))  Area of circle = πr^2       = ((πs^2 )/4)(6+4(√2)) = πa^2 (3+2(√2)) .
$${area}\:{of}\:\bigtriangleup=\:\frac{{s}^{\mathrm{2}} }{\mathrm{2}}=\:{a}^{\mathrm{2}} \:\:\left({given}\right) \\ $$$${AB}+{BC}\:=\:\frac{{s}}{\:\sqrt{\mathrm{2}}}+{r}\:=\:{r}\sqrt{\mathrm{2}} \\ $$$${r}=\frac{{s}}{\mathrm{2}−\sqrt{\mathrm{2}}}\:=\:\frac{{s}}{\mathrm{2}}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right) \\ $$$${Area}\:{of}\:{circle}\:=\:\pi{r}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:\frac{\pi{s}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\:=\:\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\:. \\ $$
Commented by TawaTawa1 last updated on 28/Mar/20
Wow, i appreciate your time sir. God bless you.
$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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