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Question Number 76855 by aliesam last updated on 31/Dec/19

Commented by aliesam last updated on 31/Dec/19

$$AB=BC=......=Fg=GA$$$$provethat$$$$thearea=\frac{a^2}{2}(\pi -7tan\left(\frac{\pi }{14}\right))$$

Commented by 67549972 last updated on 08/Feb/20

Answered by mr W last updated on 31/Dec/19

Commented by mr W last updated on 31/Dec/19

$$2\theta =\frac{2\pi }{7}$$$$\Rightarrow \theta =\frac{\pi }{7}$$$$R\cos \frac{\theta }{2}=\frac{a}{2}$$$$\Rightarrow R=\frac{a}{2\cos \frac{\pi }{14}}$$$$A_{blue}=\frac{R^2\sin 2\theta }{2}=\frac{a^2\sin \frac{2\pi }{7}}{8{\cos }^2\frac{\pi }{14}}$$$$A_{green}=\frac{a^2}{2}(\theta -\sin \theta )=\frac{a^2}{2}(\frac{\pi }{7}-\sin \frac{\pi }{7})$$$$$$$$A=7(A_{blue}+A_{green})$$$$=7[\frac{a^2\sin \frac{2\pi }{7}}{8{\cos }^2\frac{\pi }{14}}+\frac{a^2}{2}(\frac{\pi }{7}-\sin \frac{\pi }{7})]$$$$=\frac{a^2}{2}[\pi +\frac{7\sin \frac{2\pi }{7}}{4{\cos }^2\frac{\pi }{14}}-7\sin \frac{\pi }{7}]$$$$=\frac{a^2}{2}[\pi +\frac{7\sin \frac{\pi }{7}\cos \frac{\pi }{7}}{2{\cos }^2\frac{\pi }{14}}-7\sin \frac{\pi }{7}]$$$$=\frac{a^2}{2}[\pi +7\sin \frac{\pi }{7}(\frac{\cos \frac{\pi }{7}}{2{\cos }^2\frac{\pi }{14}}-1)]$$$$=\frac{a^2}{2}[\pi +7\sin \frac{\pi }{7}\left(\frac{\cos \frac{\pi }{7}-2{\cos }^2\frac{\pi }{14}}{2{\cos }^2\frac{\pi }{14}}\right)]$$$$=\frac{a^2}{2}[\pi -\frac{7\sin \frac{\pi }{7}}{2{\cos }^2\frac{\pi }{14}}]$$$$=\frac{a^2}{2}[\pi -\frac{14\sin \frac{\pi }{14}\cos \frac{\pi }{14}}{2{\cos }^2\frac{\pi }{14}}]$$$$=\frac{a^2}{2}[\pi -\frac{7\sin \frac{\pi }{14}}{\cos \frac{\pi }{14}}]$$$$=\frac{a^2}{2}(\pi -7\tan \frac{\pi }{14})$$

Commented by jagoll last updated on 31/Dec/19

$$waw....fantasticsir$$

Commented by aliesam last updated on 31/Dec/19

$$briliantsolutionsir.thankyou$$

Commented by Tawa11 last updated on 29/Dec/21

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

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