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Question-194560




Question Number 194560 by Denis last updated on 09/Jul/23
Answered by MM42 last updated on 10/Jul/23
ΔXAB  is  equilatral triangle  ∠ XAD=∠XBC=30^0    S_1 =S_2 ⇒S_1 +S_2 =(1/6)π×21^2 =11×21  S_3 =((√3)/4)×21^2   ⇒ans=21^2 −(11×21+((√3)/4)×21^2 )≈ 19
$$\Delta{XAB}\:\:{is}\:\:{equilatral}\:{triangle} \\ $$$$\angle\:{XAD}=\angle{XBC}=\mathrm{30}^{\mathrm{0}} \: \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \Rightarrow{S}_{\mathrm{1}} +{S}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\pi×\mathrm{21}^{\mathrm{2}} =\mathrm{11}×\mathrm{21} \\ $$$${S}_{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{21}^{\mathrm{2}} \\ $$$$\Rightarrow{ans}=\mathrm{21}^{\mathrm{2}} −\left(\mathrm{11}×\mathrm{21}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{21}^{\mathrm{2}} \right)\approx\:\mathrm{19} \\ $$
Commented by MM42 last updated on 10/Jul/23

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