![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2019-11-19-at-11.00.49-PM.png\")
<\/figure><\/div>\n\n\n\nTo answer the question, we need to draw line $\\overline{BE}$ and $\\overline{CE}$, as shown in the following diagram:<\/p>\n\n\n\n
![\"\"](\"https:\/\/mathfun4kids.com\/mlog\/wp-content\/uploads\/2020\/10\/Screen-Shot-2019-11-19-at-11.01.54-PM.png\")
<\/figure><\/div>\n\n\n\nObviously, $\\triangle{BCE}$ is an equalateral triangle, with its area as $\\sqrt{3}{4}$. Additionally, the size of the fan-shaped area $[CBE]$ and $[BCE]$ is $\\dfrac{\\pi}{6}$. The area of the wedge-shaped area $[CED]$ and $[BEA]$ is $\\dfrac{\\pi}{12}$. <\/p>\n\n\n\n
Therefore, the area of the region bounded by line $\\overline{CE}$ and arc $\\overparen{EC}$ is $$[CEC]=\\dfrac{\\pi}{6}-\\dfrac{\\sqrt{3}}{4}$$<\/p>\n\n\n\n
We have the area of the region bounded by $\\overline{BC}$, arc $\\overparen{BE}$ and $\\overparen{CE}$ is $$\\dfrac{\\pi}{6}+(\\dfrac{\\pi}{6}-\\dfrac{\\sqrt{3}}{4})=\\dfrac{\\pi}{3}-\\dfrac{\\sqrt{3}}{4}$$<\/p>\n\n\n\n
And the area of the region bounded by $\\overline{CD}$, arc $\\overparen{DE}$ and $\\overparen{EC}$ (and congruent region on the left side) is $$\\dfrac{\\pi}{12}-(\\dfrac{\\pi}{6}-\\dfrac{\\sqrt{3}}{4})=\\dfrac{\\sqrt{3}}{4}-\\dfrac{\\pi}{12}$$<\/p>\n\n\n\n
Finally, the area of the region bounded by line $\\overline{AD}$, arc $\\overparen{AE}$ and $\\overparen{DE}$ is $$1-(\\dfrac{\\pi}{3}-\\dfrac{\\sqrt{3}}{4})-2\\cdot(\\dfrac{\\sqrt{3}}{4}-\\dfrac{\\pi}{12})=1-\\dfrac{\\sqrt{3}}{4}-\\dfrac{\\pi}{6}$$<\/p>\n\n\n\n
<\/p>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Continuing this topic, look at the following diagram, with two quarter circles in a unit square, the question is: what are the area of of each of the 4 regions inside the square? To answer the question, we need to draw … Continue reading