Question Number 223137 by fantastic last updated on 15/Jul/25
Commented by fantastic last updated on 15/Jul/25
$${red}\:{area}?? \\ $$
Answered by fantastic last updated on 16/Jul/25
Commented by fantastic last updated on 16/Jul/25
$${R}=\mathrm{16}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{16}−{r}\right)^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{256}−\mathrm{32}{r}+\mathrm{64}=\mathrm{0} \\ $$$${r}=\mathrm{10} \\ $$$$\mathbb{A}=\frac{\mathrm{1}}{\mathrm{4}}\pi{R}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\pi\mathrm{256}×\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{100} \\ $$$$=\mathrm{128}\pi−\mathrm{50}\pi=\mathrm{78}\pi\checkmark \\ $$
Commented by fantastic last updated on 16/Jul/25
Commented by fantastic last updated on 16/Jul/25
$${D}\:{is}\:{midpoint}\:{of}\:{AC}\:{and}\:{DE}\bot{AB} \\ $$$${CBis}\:{also}\bot{on}\:{AB}\Rightarrow{E}\:{is}\:{midpoint}\:{of}\:{AB} \\ $$$${in}\bigtriangleup{ADE}\:{and}\bigtriangleup{BDE} \\ $$$${DE}\:{common},\measuredangle{DEA}=\measuredangle{DEB}=\mathrm{90}^{\mathrm{0}} \:{and}\:{AE}={EB} \\ $$$${So}\:\bigtriangleup{ADE}\cong{EDB}\left({S}−{A}−{S}\right) \\ $$$${So}\:{AD}={DB}={DC}\left(\mathscr{PROVED}\right) \\ $$