入面有1個貼邊既 圓型既4方之1
而個圓型既4方之1入面又有1個貼邊既圓型
求圓型既半徑
更新1:
4份之一
數學難題!!!!
回答 (3)
2007-12-18 5:32 pm
✔ 最佳答案
http://ustalan.googlepages.com/circle.JPG以上係sample 既圖
假設正方形既邊長係 x
咁貼邊既 4份之1圓 既半徑都會係 x, 即係AC = x
假設細圓既半徑 係r
咁 AB = OB = OC = r
而用畢氏定理. OA^2 = OB^2 + AB^2 = 2r^2
咁 AC = OA + OC = 開方(2r^2) + r = x
2r^2 = (x-r)^2
2r^2 = x^2 - 2xr + r^2
r^2 + 2xr - x^2 = 0
咁只要代返x 入去..就可以搵到個r 即係細圓既半徑出黎
希望幫到你
參考: 自己既計算
2007-12-18 11:52 pm
假設正方形既邊長係 x
貼邊既 4份之1圓 既半徑都會係 x, 即係AC = x
假設細圓既半徑 係r
AB = OB = OC = r
而用畢氏定理. OA^2 = OB^2 + AB^2 = 2r^2
AC = OA + OC = 開方(2r^2) + r = x
2r^2 = (x-r)^2
2r^2 = x^2 - 2xr + r^2
r^2 + 2xr - x^2 = 0
貼邊既 4份之1圓 既半徑都會係 x, 即係AC = x
假設細圓既半徑 係r
AB = OB = OC = r
而用畢氏定理. OA^2 = OB^2 + AB^2 = 2r^2
AC = OA + OC = 開方(2r^2) + r = x
2r^2 = (x-r)^2
2r^2 = x^2 - 2xr + r^2
r^2 + 2xr - x^2 = 0
參考: hk
2007-12-18 7:25 am
咩叫做4方之一?
收錄日期: 2021-04-23 19:12:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071217000051KK04453