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In the fig.,a circular disc with radius r cm is leaned against the wall TH.TK is the horizontal ground.A and B are the points of contact of the disc with the ground and wall respectively.TP=10cm and TR=8cm(tangent to the circle).
a) Express PR in terms of r.
b) Hence find the value of r,correct to 3 significant figures
ans:
a) 2r-18
b)15.4 cm
thx a lot!
CIRCLES..
2007-01-10 6:43 am
回答 (2)
2007-01-10 9:04 am
✔ 最佳答案
(a) Let Q be the points of contact of PR∵PR is the tangent to the circle
∴AP = PQ and BR = QR
AT = BT = r
AP = AT - TP
AP = r - 10
∴PQ = r - 10
BR = BT - TR
BR = r - 8
∴QR = r - 8
PR = PQ + QR
= (r - 10) + (r - 8)
= 2r - 18
(b) 用畢氏定理
(PR)^2 = (TP)^2 + (TR)^2
(2r - 18)^2 = (10)^2 + (8)^2
4(r^2 - 18r + 81) = 100 + 64
r^2 - 18r + 81 = 41
r^2 - 18r + 40 = 0
r = {-(-18) ±√[(-18)^2 - 4(1)(40)]} / 2(1)
= (18 ±√164)/2
= 15.4 or 2.60 (rejected)
你ans唔應該有cm兩個字
the value of r 淨係要個值
收錄日期: 2021-04-12 21:27:45
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