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所跟帖: ƽÕý ÔÙÀ´Á½µÀÌ⣺   2021-09-19 06:23:20  


作者: ¼¦Í·Èâ   ÊÔ´ð 2021-09-19 08:43:20  [点击:1416]


ڵһУԲİ뾶 \(r\)α߳ \(a\)Բĵζߵľ \(x\)ͼп
\[
r^2-x^2=(a/2)^2,\quad
r+x=a\Rightarrow\left\{\begin{array}{l}
\displaystyle
x=\frac{3r}{5}
\\
\displaystyle
a=\frac{8r}{5}
\end{array}\right.
\]
ܳıȽϣ
\[
\begin{array}{l}
\displaystyle
2\pi r-4a=2\left(\pi-\frac{16}{5}\right)r<0
\\
\displaystyle
\pi r^2-a^2=\left(\pi-\frac{64}{25}\right)r^2>0
\end{array}
\]
���༭ʱ��: 2021-09-19 08:44:20

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