初中圆的应用题

2025-01-03 21:28:40
推荐回答(5个)
回答1:

设圆半径 = r,由圆心 O 向AC做垂线,设垂足为 E,则OE = r -20,AE = 100
直角三角形 OAE中,
r^2 = (r - 20)^2 + 100^2
解得 r = 260
因为这个圆弧形门所在的圆与水平地面是相切的,所以圆弧形门的最高点离地面的高度即为圆的直径 2r =520 cm

回答2:

BD中点就是切点,设为E
圆心设为O,OE=半径r,圆弧形门的最高点离地面的高度是直径2r

连接AC,中点设为F
那么在直角三角形OAF中,OA=半径r,OF=OE-EF=半径r-20,AF=200/2=100

显然有OA的平方=OF的平方+AF的平方

代入,算出半径r=260厘米

所以圆弧形门的最高点离地面的高度是直径2r=520厘米

回答3:

AC=BD=200
设圆心O,半径R
过O作OM ⊥AC
则,AM=CM=0.5AC=100
OM=R-20
在直角三角形OAM中,由勾股定理得到:
R^2=100*100+(R-20)^2
R=260
高度=2R=520cm

回答4:

设圆弧形门的直径为R
(R/2)^2=(BD/2)^2+(R/2-AB)^2
R^2/4=10000+R^2/4-20R+400
20R=10400
R=520cm=5.2m

回答5:

设最高点离AC的高度为2r
圆心离AC的高度为h-r
因为BD是水平的,所以r²=100²+(h-r)²
h²-2rh+10000=0,h-r+20=r
所以2r(2r-20)-(2r-20)²=10000
40r=10400
r=260
h=2r=500
高度为500+20=520

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