Question-169745

Question Number 169745 by cherokeesay last updated on 07/May/22

Answered by mr W last updated on 07/May/22

Commented by mr W last updated on 08/May/22

AB=(√3) OA=OB=OD=R=((AB)/2)=((√3)/2) [ABC]=((1×(√3))/2)=((√3)/2) [AOD]=(1/2)(((√3)/2))^2 sin 120°=((3(√3))/(16)) [OBD^(⌢) ]=(π/6)(((√3)/2))^2 =(π/8) hatched area=((√3)/2)−((3(√3))/(16))−(π/8)=((5(√3)−2π)/(16))

$${AB}=\sqrt{\mathrm{3}} \\ $$$${OA}={OB}={OD}={R}=\frac{{AB}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left[{ABC}\right]=\frac{\mathrm{1}×\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\left[{AOD}\right]=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \mathrm{sin}\:\mathrm{120}°=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}} \\ $$$$\left[{O}\overset{\frown} {{BD}}\right]=\frac{\pi}{\mathrm{6}}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\pi}{\mathrm{8}} \\ $$$${hatched}\:{area}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}}−\frac{\pi}{\mathrm{8}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{2}\pi}{\mathrm{16}} \\ $$

Commented by cherokeesay last updated on 07/May/22

$${very}\:{nice}\:! \\ $$$${thank}\:{you}\:{sir}\:! \\ $$

Commented by Tawa11 last updated on 08/Oct/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by som(math1967) last updated on 07/May/22

AB=BCtan60=(√3)cm OB=((√3)/2)cm ∡BOL=2×30=60 OB=OL⇒△OBL equilateral ∠LBC=90−60=30 ar. of △OBL=((√3)/4)×(((√3)/2))^2 =((3(√3))/(16))cm^2 ar.of △BLC=(1/2)×1×((√3)/2)sin∠LBC =((√3)/8)cm^2 ar.of sectorBOL=(1/6)π×(((√3)/2))^2 cm^2 (π/8)cm^2 hatched area=(((3(√3))/(16)) +((√3)/8))−(π/8) =(((5(√3))/(16)) −(π/8))cm^2

$${AB}={BCtan}\mathrm{60}=\sqrt{\mathrm{3}}{cm} \\ $$$${OB}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cm} \\ $$$$\measuredangle{BOL}=\mathrm{2}×\mathrm{30}=\mathrm{60} \\ $$$${OB}={OL}\Rightarrow\bigtriangleup{OBL}\:{equilateral} \\ $$$$\angle{LBC}=\mathrm{90}−\mathrm{60}=\mathrm{30} \\ $$$${ar}.\:{of}\:\bigtriangleup{OBL}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}}{cm}^{\mathrm{2}} \\ $$$${ar}.{of}\:\bigtriangleup{BLC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\angle{LBC} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}{cm}^{\mathrm{2}} \\ $$$${ar}.{of}\:{sectorBOL}=\frac{\mathrm{1}}{\mathrm{6}}\pi×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} {cm}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\pi}{\mathrm{8}}{cm}^{\mathrm{2}} \\ $$$${hatched}\:{area}=\left(\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{8}} \\ $$$$=\left(\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{16}}\:−\frac{\pi}{\mathrm{8}}\right){cm}^{\mathrm{2}} \\ $$

Commented by som(math1967) last updated on 07/May/22