<例題>下の図の半円の弦 BC、CD、DA の長さが 6,6,2.8 のとき、この円の半径を
求めよ。
<解答>半円の半径を r とする。
cos(∠ABC)=6/2r=3/r ∵ ∠BCA=π/2
cos(∠ADC)=cos(π−∠ABC)=−cos(∠ABC)=−3/r
条件より、(AC)S=(AD+DC)S=(AD)S+2(AD・DC)+(DC)S
=(AD)S−2(DA・DC)+(DC)S
=(2.8)2−2×2.8×6×cos(∠ADC)+(6)2
=(2.8)2−2×2.8×6×(−3/r)+(6)2
=43.84+100.8/r
43.84+100.8/r=(AC)S
=(AB+BC)S
=(AB)S+2(AB・BC)+(BC)S
=(AB)S+2{(AC+CB)・BC}+(BC)S
=(AB)S+2(AC・BC)−2(BC)S+(BC)S
=(AB)S+2(0)−2(BC)S+(BC)S
=(AB)S−(BC)S
=(2r)2−(6)2
=4r2−36
0=4r2−36−43.84−100.8/r
=4r3−36r−43.84r−100.8
=4r3−79.84r−100.8
=r3−19.96r−25.2
=(r−5)(r2+5r+5.04)
=(r−5) ∵ (r2+5r+5.04)≠0
r=5・・・・・・・・・・・・・・・・・・(答)
|