Question Number 194560 by Denis last updated on 09/Jul/23
Answered by MM42 last updated on 10/Jul/23
$$\Delta{XAB}\:\:{is}\:\:{equilatral}\:{triangle} \\ $$$$\angle\:{XAD}=\angle{XBC}=\mathrm{30}^{\mathrm{0}} \: \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\pi×\mathrm{21}^{\mathrm{2}} =\mathrm{11}×\mathrm{21} \\ $$$${S}_{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{21}^{\mathrm{2}} \\ $$$$\Rightarrow{ans}=\mathrm{21}^{\mathrm{2}} −\left(\mathrm{11}×\mathrm{21}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{21}^{\mathrm{2}} \right)\approx\:\mathrm{19} \\ $$
Commented by MM42 last updated on 10/Jul/23
