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Question Number 218733 by Spillover last updated on 14/Apr/25

Commented by A5T last updated on 15/Apr/25

$$\mathrm{This}\mathrm{leads}\mathrm{to}\mathrm{a}\mathrm{contradiction}(\mathrm{if}\mathrm{the}\mathrm{other}\mathrm{chord}$$$$\mathrm{were}\mathrm{also}\mathrm{a}\mathrm{diameter}).$$

Answered by mr W last updated on 15/Apr/25

Commented by mr W last updated on 15/Apr/25

$$R=\sqrt{3^2+{\left(\frac{3}{2}\right)}^2}=\frac{3\sqrt{5}}{2}$$$$a=R-\frac{3}{2}=\frac{3(\sqrt{5}-1)}{2}$$$$b=2R-3a=2\times \frac{3\sqrt{5}}{2}-3\times \frac{3(\sqrt{5}-1)}{2}=\frac{3(3-\sqrt{5})}{2}$$$$c^2=3^2+b^2=\frac{27(3-\sqrt{5})}{2}$$$$cd=2a(a+b)$$$$\frac{h}{3}=\frac{d}{c}=\frac{2a(a+b)}{c^2}$$$$=\frac{2\times 2}{27(3-\sqrt{5})}\times \frac{3(\sqrt{5}-1)}{2}\times (\frac{3(\sqrt{5}-1)}{2}+\frac{3(3-\sqrt{5})}{2})$$$$=\frac{2(\sqrt{5}-1)}{3(3-\sqrt{5})}=\frac{\sqrt{5}+1}{3}$$$$\Rightarrow h=\sqrt{5}+1$$$$A_{pink}=\frac{2ah}{2}=ah=\frac{3(\sqrt{5}-1)(\sqrt{5}+1)}{2}=6✓$$

Commented by Spillover last updated on 15/Apr/25

$$verynice.thanks$$

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

Answered by Spillover last updated on 15/Apr/25

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