Question Number 60812 by peter frank last updated on 26/May/19
Commented by tanmay last updated on 26/May/19
![area of BCADB=2×(1/2)×5×12=60 60=2[(1/2)×13×h] h=((60)/(13)) sinθ_1 =(h/(12)) and sinθ_2 =(h/5) sinθ_1 =((60)/(13×12))=(5/(13)) sinθ_2 =((12)/(13)) area of triangle △BCD=(1/2)×CD×BO =(1/2)×2h×12cosθ_1 [cosθ_1 =((BO)/(12))] =((60)/(13))×12×((12)/(13))=60×(((12)/(13)))^2 area of trisngle △CAD=(1/2)×CD×AO =(1/2)×2h×5cosθ_2 [cosθ_2 =((AO)/5)] =((60)/(13))×5×(5/(13))=60×((5/(13)))^2 area of sector=((πr^2 )/(2π))×θ=(1/2)r^2 θ here one sector area=(1/2)×12^2 ×2θ_1 = another sector area=(1/2)×5^2 ×2θ_2 shaded area=[12^2 θ_1 +5^2 θ_2 ]−[60×(((12)/(13)))^2 +60×((5/(13)))^2 ] =144sin^(−1) ((5/(13)))+25sin^(−1) (((12)/(13)))−60](Q60821.png)
$${area}\:{of}\:{BCADB}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}×\mathrm{12}=\mathrm{60} \\ $$$$\mathrm{60}=\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{13}×{h}\right] \\ $$$${h}=\frac{\mathrm{60}}{\mathrm{13}} \\ $$$${sin}\theta_{\mathrm{1}} =\frac{{h}}{\mathrm{12}}\:\:\:{and}\:{sin}\theta_{\mathrm{2}} =\frac{{h}}{\mathrm{5}} \\ $$$${sin}\theta_{\mathrm{1}} =\frac{\mathrm{60}}{\mathrm{13}×\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{13}}\:\:\:{sin}\theta_{\mathrm{2}} =\frac{\mathrm{12}}{\mathrm{13}} \\ $$$${area}\:{of}\:{triangle}\:\bigtriangleup{BCD}=\frac{\mathrm{1}}{\mathrm{2}}×{CD}×{BO} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{h}×\mathrm{12}{cos}\theta_{\mathrm{1}} \:\:\:\left[{cos}\theta_{\mathrm{1}} =\frac{{BO}}{\mathrm{12}}\right] \\ $$$$=\frac{\mathrm{60}}{\mathrm{13}}×\mathrm{12}×\frac{\mathrm{12}}{\mathrm{13}}=\mathrm{60}×\left(\frac{\mathrm{12}}{\mathrm{13}}\right)^{\mathrm{2}} \\ $$$${area}\:{of}\:{trisngle}\:\bigtriangleup{CAD}=\frac{\mathrm{1}}{\mathrm{2}}×{CD}×{AO} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{h}×\mathrm{5}{cos}\theta_{\mathrm{2}} \:\:\left[{cos}\theta_{\mathrm{2}} =\frac{{AO}}{\mathrm{5}}\right] \\ $$$$=\frac{\mathrm{60}}{\mathrm{13}}×\mathrm{5}×\frac{\mathrm{5}}{\mathrm{13}}=\mathrm{60}×\left(\frac{\mathrm{5}}{\mathrm{13}}\right)^{\mathrm{2}} \\ $$$${area}\:{of}\:{sector}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}\pi}×\theta=\frac{\mathrm{1}}{\mathrm{2}}{r}^{\mathrm{2}} \theta \\ $$$${here}\:{one}\:{sector}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}^{\mathrm{2}} ×\mathrm{2}\theta_{\mathrm{1}} = \\ $$$${another}\:{sector}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{5}^{\mathrm{2}} ×\mathrm{2}\theta_{\mathrm{2}} \\ $$$${shaded}\:{area}=\left[\mathrm{12}^{\mathrm{2}} \theta_{\mathrm{1}} +\mathrm{5}^{\mathrm{2}} \theta_{\mathrm{2}} \right]−\left[\mathrm{60}×\left(\frac{\mathrm{12}}{\mathrm{13}}\right)^{\mathrm{2}} +\mathrm{60}×\left(\frac{\mathrm{5}}{\mathrm{13}}\right)^{\mathrm{2}} \right] \\ $$$$=\mathrm{144}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{5}}{\mathrm{13}}\right)+\mathrm{25}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{12}}{\mathrm{13}}\right)−\mathrm{60} \\ $$$$ \\ $$
Commented by peter frank last updated on 26/May/19
$${find}\:{area}\:{of}\:{shaded}\:{region} \\ $$
Commented by tanmay last updated on 26/May/19
Commented by peter frank last updated on 26/May/19
$${thank}\:{you}\:{sir} \\ $$
Commented by tanmay last updated on 26/May/19
$${most}\:{welcme}... \\ $$