Question Number 207341 by mr W last updated on 12/May/24
Answered by som(math1967) last updated on 12/May/24
Commented by som(math1967) last updated on 12/May/24
$${ar}\:{of}\:\bigtriangleup{BOC} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{7}×\mathrm{7}×{sin}\mathrm{120}=\frac{\mathrm{49}\sqrt{\mathrm{3}}}{\mathrm{4}}\:{squnit} \\ $$$$\:{BC}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +\mathrm{2}×\mathrm{98} \\ $$$${BC}=\mathrm{7}\sqrt{\mathrm{3}} \\ $$$${of}\:\bigtriangleup{ABC}\:\therefore\angle{BAC}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{11}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} −\mathrm{49}×\mathrm{3}}{\mathrm{2}×\mathrm{11}×\mathrm{13}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{60} \\ $$$${ar}.\:\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{11}×\mathrm{13}{sin}\mathrm{60} \\ $$$$\:\:=\frac{\mathrm{143}\sqrt{\mathrm{3}}}{\mathrm{4}}{sq}\:{unit} \\ $$$$\:{Yellow}\:{area}=\frac{\mathrm{143}\sqrt{\mathrm{3}}}{\mathrm{4}}\:−\frac{\mathrm{49}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{47}\sqrt{\mathrm{3}}}{\mathrm{2}}{squnit} \\ $$$$ \\ $$
Commented by som(math1967) last updated on 12/May/24
$$\:{yes}\:{it}\:{may}\:{not}\:{circumcentre} \\ $$
Commented by mr W last updated on 12/May/24
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Answered by mr W last updated on 12/May/24
$$\mathrm{11}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} −\mathrm{2}×\mathrm{11}×\mathrm{13}\:\mathrm{cos}\:\alpha=\mathrm{7}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{2}×\mathrm{7}×\mathrm{7}\:\mathrm{cos}\:\mathrm{120}° \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\alpha=\mathrm{60}° \\ $$$${A}_{{yellow}} =\frac{\mathrm{11}×\mathrm{13}×\mathrm{sin}\:\mathrm{60}°}{\mathrm{2}}−\frac{\mathrm{7}×\mathrm{7}×\mathrm{sin}\:\mathrm{120}°}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{47}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$