{"id":367,"date":"2019-12-02T01:25:05","date_gmt":"2019-12-02T01:25:05","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=367"},"modified":"2020-10-15T06:34:39","modified_gmt":"2020-10-15T06:34:39","slug":"circles-in-a-square-part-6","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=367","title":{"rendered":"Circles in a Square – Part 6"},"content":{"rendered":"\n
Look at the following figure, a simi-circle is inscribed between the quarter circle and one side of the unit square. What the radius of the semi-circle?<\/p>\n\n\n\n
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Obviously, $G$ is the center of the semi-circle, line $\\overline{CG}$ crosses $F$, the tangent point of the quarter circle and semi-circle, and $\\triangle{BCG}$ is a right triangle. According to Pythagoras theorem, we have\n$$\\overline{BC}^2 + \\overline{BG}^2 = \\overline{CG}^2$$\nAssume the radius of the semi-circle is $r$, we have\n$$\\overline{BC}=1$$\n$$\\overline{BG}=\\overline{AB}-\\overline{AG}=1-r$$\n$$\\overline{CG}=\\overline{CF}+\\overline{GF}=1+r$$\nTherefore\n$$1^2+(1-r)^2=(1+r)^2$$\nThe above equation can be simplified as $$4r=1$$\nTherefore the radius of the semi-circle is $\\dfrac{1}{4}$.\n<\/p>\n\n\n\n
Can you solve the problem without using Pythagoras theorem? To be continued…<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"
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