Picture: http://postimg.org/image/sniez3orj/
In the figure, two equal circles with centre H and K respectively intersect at points A and B. AB intersects HK at N. HPNQK is a straight line. If AB=HK=10 cm,
find PQ
The above question is similar to question 1:
https://hk.knowledge.yahoo.com/question/question?qid=7014111700003
But AN=NB is not provided in this question, how can I find it or using another
method to do this question?
Thank you!
F.5 maths circle
2014-11-24 1:35 am
回答 (3)
2014-11-24 2:57 am
✔ 最佳答案
JoinAH, AK, BH and BK.
AH = AK = BH = BK (radii of equal circles, radii of same circle)
Hence, AHBK is a rhombus.
AN = BN = (1/2)AB = 5 cm (diagonals of rhombus bisect each other)
HN = KN = (1/2)HK = 5 cm (diagonals of rhombus bisect each other)
AN⊥KN (diagonals of rhombus⊥eachother)
In rt.∠-ΔANK :
AK² = AN² +NK² (Pythagorean theorem)
AK² = (5 cm)² + (5cm)²
AK = 5√2 cm
Radius of each circle = 5√2 cm
HPNQK is a st. line.
HK = HQ + PK - PQ
Hence, PQ = HQ + PK - HK
PQ = (5 cm) + (5 cm) - (5√2) cm
PQ = (10 - 5√2) cm
2014-11-23 18:58:44 補充：
The above is one of the methods to solve the problem.
2014-11-26 04:47:03 補充：
Amendment :
The last two lines should be :
PQ = (5√2 cm) + (5√2 cm) - (10) cm
PQ = (10√2 - 10) cm ...... (Ans)
2014-11-24 10:22 pm
PQ = HQ + PK - HK
= (5√2 cm) + (5√2 cm) - (10) cm
= (10√2－10) cm
= (5√2 cm) + (5√2 cm) - (10) cm
= (10√2－10) cm
2014-11-24 1:42 am
By symmetry, AN = NB.
I think saying in this way is acceptable.
If some teachers are more stubborn, then you claim AKBH is a rhombus and use its properties.
I think saying in this way is acceptable.
If some teachers are more stubborn, then you claim AKBH is a rhombus and use its properties.
收錄日期: 2021-04-15 17:17:17
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20141123000051KK00068