標題:
f.5 circles 1條
發問:
http://imageshack.us/photo/my-images/32/img1623v.jpg/ 題中既27題 請師兄教教me!! thz!
最佳解答:
a) i) With ∠POR = ∠OQP = 90 (∠s in semi-circle), ∠OQR = 90 since OQ is perp. to PR ∠QSO = 90 since QS//RO Then suppose ∠OQS = θ, we have ∠QOS = 90 - θ and ∠QOR = θ With ∠QSO = ∠OQR and ∠QOS = ORQ, we have △QSO ~ △OQR (AAA) So QS/OQ = QO/OR OQ2 = OR x QS ii) With ∠QPO = 180 - 90 - (90 - θ) = θ, we have ∠OQS = ∠OPQ and ∠OSQ = OQP Hence △OQS ~ △OPQ (AAA). b) i) OQ2 = 22 + 42 = 20 With OS = 2 since S is at (2, 0), we have: 2OR = 20 OR = 10 So R is at (0, 10) ii) With △OQS ~ △ROP: QS/OS = OP/OR OP = 10 x 4/2 = 20 Hence P is at (20, 0) Thus centre of circle OPR is at (10, 5) and radius = (1/2)√(102 + 202) = 5√5
其他解答:
b(i) Let OP=p, OR =r, OQ=root(20) Consider OQS and OPQ root(20) / p = 2 / root(20) p=10 4/r = (p-2)/p r =5 Coordinates of R (0, 5) PR is diameter radius = root(r^2 + p^2) / 2 = 5 x root(5) / 2 Hence, center is mid point of P and R, so is (5, 2.5)
f.5 circles 1條
發問:
http://imageshack.us/photo/my-images/32/img1623v.jpg/ 題中既27題 請師兄教教me!! thz!
最佳解答:
a) i) With ∠POR = ∠OQP = 90 (∠s in semi-circle), ∠OQR = 90 since OQ is perp. to PR ∠QSO = 90 since QS//RO Then suppose ∠OQS = θ, we have ∠QOS = 90 - θ and ∠QOR = θ With ∠QSO = ∠OQR and ∠QOS = ORQ, we have △QSO ~ △OQR (AAA) So QS/OQ = QO/OR OQ2 = OR x QS ii) With ∠QPO = 180 - 90 - (90 - θ) = θ, we have ∠OQS = ∠OPQ and ∠OSQ = OQP Hence △OQS ~ △OPQ (AAA). b) i) OQ2 = 22 + 42 = 20 With OS = 2 since S is at (2, 0), we have: 2OR = 20 OR = 10 So R is at (0, 10) ii) With △OQS ~ △ROP: QS/OS = OP/OR OP = 10 x 4/2 = 20 Hence P is at (20, 0) Thus centre of circle OPR is at (10, 5) and radius = (1/2)√(102 + 202) = 5√5
其他解答:
b(i) Let OP=p, OR =r, OQ=root(20) Consider OQS and OPQ root(20) / p = 2 / root(20) p=10 4/r = (p-2)/p r =5 Coordinates of R (0, 5) PR is diameter radius = root(r^2 + p^2) / 2 = 5 x root(5) / 2 Hence, center is mid point of P and R, so is (5, 2.5)
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