# 问题及解答

## 求下列图形中绿色部分的面积.

Posted by haifeng on 2017-05-08 23:40:16 last update 2017-05-12 09:44:08 | Edit | Answers (2)

$25\arccos\frac{1}{2\sqrt{2}}-100\arccos\frac{5}{4\sqrt{2}}+\frac{25}{2}\sqrt{7}$

1

Posted by haifeng on 2017-05-09 09:30:06

$\cos\theta=\frac{|OA|^2+|AB|^2-|OB|^2}{2|OA|\cdot|AB|}=\frac{(\sqrt{2}r)^2+(2r)^2-r^2}{2\sqrt{2}r\cdot 2r}=\frac{5}{4\sqrt{2}},$

$\cos(\pi-\alpha)=\frac{|OA|^2+|OB|^2-|AB|^2}{2|OA|\cdot|OB|}=\frac{(\sqrt{2}r)^2+r^2-(2r)^2}{2\sqrt{2}r\cdot r}=-\frac{1}{2\sqrt{2}}.$

$x=\text{圆O中扇形 OBC 的面积}-S_1-S_2$

\begin{aligned} S_1&=\pi(2r)^2\cdot\frac{2\theta}{2\pi}-2r\sin\theta\cdot 2r\cos\theta=4r^2\theta-4r^2\sin\theta\cos\theta,\\ S_2&=r\sin\alpha\cdot r\cos\alpha=r^2\sin\alpha\cos\alpha \end{aligned}

$\begin{split} x&=\pi r^2\cdot\frac{2\alpha}{2\pi}-(4r^2\theta-4r^2\sin\theta\cos\theta)-r^2\sin\alpha\cos\alpha\\ &=(\alpha -4\theta+4\sin\theta\cos\theta-\sin\alpha\cos\alpha)r^2\\ &=(\arccos\frac{1}{2\sqrt{2}}-4\arccos\frac{5}{4\sqrt{2}}+4\cdot\frac{\sqrt{7}}{4\sqrt{2}}\cdot\frac{5}{4\sqrt{2}}-\frac{\sqrt{7}}{2\sqrt{2}}\cdot\frac{1}{2\sqrt{2}})r^2\\ &=(\arccos\frac{1}{2\sqrt{2}}-4\arccos\frac{5}{4\sqrt{2}}+\frac{\sqrt{7}}{2})r^2 \end{split}$

$x=25\arccos\frac{1}{2\sqrt{2}}-100\arccos\frac{5}{4\sqrt{2}}+\frac{25\sqrt{7}}{2}.$

2

Posted by haifeng on 2017-05-12 09:08:42