Question Number 90272 by student work last updated on 22/Apr/20
Answered by behi83417@gmail.com last updated on 22/Apr/20
![AO+OC=8 AO^2 +OC^2 =6^2 ⇒AO.OC=(1/2)[(AO+OC)^2 −(AO^2 +OC^2 )]= =(1/2)[64−36]=14⇒z^2 −8z+14=0 z=((8±(√(64−56)))/2)=4±(√2) ⇒ { ((AO=4+(√2))),((OC=4−(√2))) :} S_(blue) =((πr^2 )/4)−(1/2)AO.OC=((π×36)/4)−(1/2)×14= =9π−7 P_(blue) =ST+CT+CA+AS= =((πr)/2)+[6−(4−(√2))]+(6)+[(6−(4+(√2))]= =3π+10 .■](https://www.tinkutara.com/question/Q90283.png)
$$\mathrm{AO}+\mathrm{OC}=\mathrm{8} \\ $$$$\mathrm{AO}^{\mathrm{2}} +\mathrm{OC}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{AO}.\mathrm{OC}=\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\mathrm{AO}+\mathrm{OC}\right)^{\mathrm{2}} −\left(\mathrm{AO}^{\mathrm{2}} +\mathrm{OC}^{\mathrm{2}} \right)\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{64}−\mathrm{36}\right]=\mathrm{14}\Rightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{8z}+\mathrm{14}=\mathrm{0} \\ $$$$\mathrm{z}=\frac{\mathrm{8}\pm\sqrt{\mathrm{64}−\mathrm{56}}}{\mathrm{2}}=\mathrm{4}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\mathrm{AO}=\mathrm{4}+\sqrt{\mathrm{2}}}\\{\mathrm{OC}=\mathrm{4}−\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{S}_{\mathrm{blue}} =\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AO}.\mathrm{OC}=\frac{\pi×\mathrm{36}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{14}= \\ $$$$=\mathrm{9}\pi−\mathrm{7} \\ $$$$\mathrm{P}_{\mathrm{blue}} =\mathrm{ST}+\mathrm{CT}+\mathrm{CA}+\mathrm{AS}= \\ $$$$=\frac{\pi\mathrm{r}}{\mathrm{2}}+\left[\mathrm{6}−\left(\mathrm{4}−\sqrt{\mathrm{2}}\right)\right]+\left(\mathrm{6}\right)+\left[\left(\mathrm{6}−\left(\mathrm{4}+\sqrt{\mathrm{2}}\right)\right]=\right. \\ $$$$=\mathrm{3}\pi+\mathrm{10}\:\:.\blacksquare \\ $$
Commented by student work last updated on 22/Apr/20
$$\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{\pi×\mathrm{36}}{\mathrm{4}}\:\:\:\:{you}\:{wrong}? \\ $$
Commented by ajfour last updated on 22/Apr/20
$$\mathrm{S}_{\mathrm{blue}} =\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AO}.\mathrm{OC}=\frac{\pi×\mathrm{36}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{14}= \\ $$$$\:\:\:=\:\mathrm{9}\pi−\mathrm{7} \\ $$