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
Chords of a circle f4
2011-04-17 1:55 am
回答 (3)
2011-04-17 2:29 am
✔ 最佳答案
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 
2011-04-17 2:36 am
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
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
2011-04-17 2:30 am
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。
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。
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