Question Number 197719 by a.lgnaoui last updated on 27/Sep/23
$$\mathrm{Determiner}\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$
Commented by a.lgnaoui last updated on 27/Sep/23
Commented by som(math1967) last updated on 27/Sep/23
Commented by som(math1967) last updated on 27/Sep/23
$${let}\:{rad}\:{of}\:{circle}={r} \\ $$$${O}\:{is}\:{centre}\:{of}\:{circle} \\ $$$${ABCD}\:{rectangle}\:{O}\:{is}\:{point}\: \\ $$$${in}\:{the}\:{rectangle} \\ $$$$\therefore{OB}^{\mathrm{2}} +{OD}^{\mathrm{2}} ={OC}^{\mathrm{2}} +{OA}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{7}^{\mathrm{2}} +{r}^{\mathrm{2}} +{x}^{\mathrm{2}} +{r}^{\mathrm{2}} =\mathrm{17}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{11}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{17}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} +\mathrm{121} \\ $$$$\therefore{x}=\sqrt{\mathrm{361}}=\mathrm{19} \\ $$$$\left.\:{b}\right)\:\mathrm{19} \\ $$