Question Number 198182 by Blackpanther last updated on 13/Oct/23
Answered by mr W last updated on 14/Oct/23
Commented by mr W last updated on 14/Oct/23
$${r}={radius}\:{of}\:{circles}=\mathrm{1} \\ $$$$\angle{QPS}=\mathrm{60}°=\angle{QRP} \\ $$$$\angle{PSQ}=\angle{QRP}/\mathrm{2}=\mathrm{30}° \\ $$$$\Rightarrow\Delta{PQS}={right}\:{angled} \\ $$$$\Delta{ABC}\sim\Delta{QPS} \\ $$$${AB}=\mathrm{3}{r}+\sqrt{\mathrm{3}}{r}=\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){r} \\ $$$$\Delta{ABC}=\frac{{AB}×{AC}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\frac{\left({AB}\right)^{\mathrm{2}} ×\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \sqrt{\mathrm{3}}{r}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{9}+\mathrm{6}\sqrt{\mathrm{3}}\right){r}^{\mathrm{2}} =\mathrm{9}+\mathrm{6}\sqrt{\mathrm{3}}\:\:\:\checkmark \\ $$