Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

$$arof△BOC$$$$\frac{1}{2}\times 7\times 7\times sin120=\frac{49\sqrt{3}}{4}squnit$$$${BC}^2=7^2+7^2+2\times 98$$$$BC=7\sqrt{3}$$$$of△ABC\therefore \mathrm{\angle }BAC={\cos }^{-1}\left(\frac{{11}^2+{13}^2-49\times 3}{2\times 11\times 13}\right)$$$$=60$$$$ar.△ABC=\frac{1}{2}\times 11\times 13sin60$$$$=\frac{143\sqrt{3}}{4}squnit$$$$Yellowarea=\frac{143\sqrt{3}}{4}-\frac{49\sqrt{3}}{4}$$$$=\frac{47\sqrt{3}}{2}squnit$$$$$$

Commented by som(math1967) last updated on 12/May/24

$$yesitmaynotcircumcentre$$

Commented by mr W last updated on 12/May/24

��

Answered by mr W last updated on 12/May/24

$${11}^2+{13}^2-2\times 11\times 13\cos \alpha =7^2+7^2-2\times 7\times 7\cos 120°$$$$\Rightarrow \cos \alpha =\frac{1}{2}\Rightarrow \alpha =60°$$$$A_{yellow}=\frac{11\times 13\times \sin 60°}{2}-\frac{7\times 7\times \sin 120°}{2}$$$$=\frac{47\sqrt{3}}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com