標題:
Chords of a circle f4
發問:
In the figure, O is the centre and the radius of the circle is 32.5cm. AB and CD intersect at K. If AB is perpendicular to CD, AB=33cm and CD=63cm, find OK. (Leave your answer in surd form.) 圖片參考：http://imgcld.yimg.com/8/n/HA06575449/o/701104160087713873427370.jpg
最佳解答:
Let x = The distance between centre O to line AB y = The distance between centre O to line CD r = radius of circle OK^2 = x^2 + y^2r^2 = x^2 + (33/2)^2r^2 = y^2 + (63/2)^2 OK^2 = [r^2-(33/2)^2]+[r^2-(63/2)^2]OK^2 = 2(32.5)^2-(16.5)^2-(31.5)^2OK^2 = 848OK = √848OK =4√53
其他解答:
此文章來自奇摩知識+如有不便請留言告知
Sol 過O作一直線垂直AB交AB於E 過O作一直線垂直CD交CD於F OE^2=32.5^2-(33/2)^2=784 OE=28 OF^2=32.5^2-(63/2)^2=64 OF=8 OK^2=8^2+28^2=64+784=848 OK=4√53|||||O至AB的垂直距離 = √(32.5^2 - 16.5^2) = 28 O至CD的垂直距離 = √(32.5^2 - 31.5^2) = 8 因此OK = √(28^2 + 8^2) = √848 = 4√53 cm 2011-04-16 18:32:07 補充： 幅圖晝得不太好。因為圓心至弦的垂線是平分弦的。
留言列表
