Question-93125

Question Number 93125 by john santu last updated on 11/May/20

Commented by john santu last updated on 11/May/20

$$\mathrm{find}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{area}\: \\ $$

Commented by i jagooll last updated on 11/May/20

(r/6) = (x/(x+2r))⇒ rx+2r^2 =6x 2r^2 = x(6−r) ⇒x = ((2r^2 )/(6−r)) ...(i) 6^2 = r(2r+x) ⇒36=r(2r+((2r^2 )/(6−r)))..(ii) 18 =r( r+(r^2 /(6−r)) )⇒18=r(((6r)/(6−r))) 18−3r = r^2 ⇒r^2 +3r−18=0 (r+6)(r−3)=0 ⇒r=3 shaded area = πr^2 = 9π

$$\frac{\mathrm{r}}{\mathrm{6}}\:=\:\frac{\mathrm{x}}{\mathrm{x}+\mathrm{2r}}\Rightarrow\:\mathrm{rx}+\mathrm{2r}^{\mathrm{2}} =\mathrm{6x} \\ $$$$\mathrm{2r}^{\mathrm{2}} \:=\:\mathrm{x}\left(\mathrm{6}−\mathrm{r}\right)\:\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{2r}^{\mathrm{2}} }{\mathrm{6}−\mathrm{r}}\:…\left(\mathrm{i}\right) \\ $$$$\mathrm{6}^{\mathrm{2}} \:=\:\mathrm{r}\left(\mathrm{2r}+\mathrm{x}\right)\:\Rightarrow\mathrm{36}=\mathrm{r}\left(\mathrm{2r}+\frac{\mathrm{2r}^{\mathrm{2}} }{\mathrm{6}−\mathrm{r}}\right)..\left(\mathrm{ii}\right) \\ $$$$\mathrm{18}\:=\mathrm{r}\left(\:\mathrm{r}+\frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{6}−\mathrm{r}}\:\right)\Rightarrow\mathrm{18}=\mathrm{r}\left(\frac{\mathrm{6r}}{\mathrm{6}−\mathrm{r}}\right) \\ $$$$\mathrm{18}−\mathrm{3r}\:=\:\mathrm{r}^{\mathrm{2}} \:\Rightarrow\mathrm{r}^{\mathrm{2}} +\mathrm{3r}−\mathrm{18}=\mathrm{0} \\ $$$$\left(\mathrm{r}+\mathrm{6}\right)\left(\mathrm{r}−\mathrm{3}\right)=\mathrm{0}\:\Rightarrow\mathrm{r}=\mathrm{3} \\ $$$$\mathrm{shaded}\:\mathrm{area}\:=\:\pi\mathrm{r}^{\mathrm{2}} \:=\:\mathrm{9}\pi\: \\ $$

Commented by john santu last updated on 11/May/20

cool man ��

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