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Question Number 183747 by HeferH last updated on 29/Dec/22

Find the perimeter of a regular heptagon ABCDEFG if (1/(AE)) + (1/(AC)) = (1/5)

$$\:{Find}\:{the}\:{perimeter}\:{of}\:{a}\:{regular}\:{heptagon}\: \\ $$$$\:{ABCDEFG}\:{if}\:\frac{\mathrm{1}}{{AE}}\:+\:\frac{\mathrm{1}}{{AC}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\: \\ $$

Answered by mr W last updated on 29/Dec/22

θ=((5π)/7) AE=2a sin (θ/2)=2a sin ((5π)/(14)) AE=a+2a sin (θ−(π/2))=a(1+2 sin ((3π)/(14))) (1/(2a sin ((5π)/(14))))+(1/(a(1+2 sin ((3π)/(14)))))=(1/5) (a/5)=(1/(2 sin ((5π)/(14))))+(1/(1+2 sin ((3π)/(14))))=1 ⇒a=5 P=7a=35

$$\theta=\frac{\mathrm{5}\pi}{\mathrm{7}} \\ $$$${AE}=\mathrm{2}{a}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{a}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$${AE}={a}+\mathrm{2}{a}\:\mathrm{sin}\:\left(\theta−\frac{\pi}{\mathrm{2}}\right)={a}\left(\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{a}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}}+\frac{\mathrm{1}}{{a}\left(\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\right)}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\frac{{a}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{14}}}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}}=\mathrm{1} \\ $$$$\Rightarrow{a}=\mathrm{5} \\ $$$${P}=\mathrm{7}{a}=\mathrm{35} \\ $$

Commented by HeferH last updated on 29/Dec/22

thanks :)

$$\left.{thanks}\::\right) \\ $$

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