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Question Number 218833 by Spillover last updated on 16/Apr/25

Answered by mr W last updated on 16/Apr/25

Commented by mr W last updated on 16/Apr/25

$$R^2=a^2+b^2$$$$\frac{b}{4}=\frac{a}{\sqrt{a^2-2^2}}\Rightarrow \frac{b^2}{16}=\frac{a^2}{a^2-2^2}$$$${(1+4)}^2=(R+\sqrt{b^2-4^2})(R-\sqrt{b^2-4^2})$$$$5^2=R^2-b^2+4^2$$$$5^2=a^2+4^2\Rightarrow a^2=3^2$$$$\Rightarrow b^2=\frac{3^2\times 16}{3^2-2^2}=\frac{144}{5}$$$${(x+2)}^2=(R+\sqrt{a^2-2^2})(R-\sqrt{a^2-2^2})$$$${(x+2)}^2=R^2-a^2+2^2=b^2+2^2$$$$\Rightarrow x=\sqrt{b^2+2^2}-2$$$$=\sqrt{\frac{144}{5}+4}-2=2(\frac{\sqrt{205}}{5}-1)\approx 5.727$$

Commented by Spillover last updated on 16/Apr/25

$$greatwork.thanks$$

Answered by Spillover last updated on 16/Apr/25

Answered by Spillover last updated on 16/Apr/25

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