Question Number 142604 by Engr_Jidda last updated on 02/Jun/21
Answered by mr W last updated on 02/Jun/21
$${a}=\mathrm{14} \\ $$$${r}_{\mathrm{1}} ={b}=\mathrm{10}.\mathrm{5} \\ $$$${r}_{\mathrm{2}} =\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }−{b} \\ $$$${r}_{\mathrm{3}} ={b}−{r}_{\mathrm{2}} =\mathrm{2}{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\alpha=\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}} \\ $$$${S}=\frac{{ab}}{\mathrm{2}}−\frac{{r}_{\mathrm{1}} ^{\mathrm{2}} \alpha}{\mathrm{2}}−\frac{{r}_{\mathrm{2}} ^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\alpha\right)}{\mathrm{2}}−\frac{\pi{r}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{4}} \\ $$
Commented by Engr_Jidda last updated on 03/Jun/21
$${thank}\:{you}\:{sir} \\ $$