F5 Maths

2010-11-30 2:36 am

圖片參考:http://imgcld.yimg.com/8/n/HA00962128/o/701011290092613873418250.jpg



( Leave the answers in terms of pie or in surd form if necessary)


In the figure, an equilateral triangle P1Q1R1 is inscribed in a circle C1. C2 is the inscribed circle of triangle P2Q2R2 is
inscribed in C2. The process is repeated indefinitely. Suppose
the radius of C1 is 8 cm.

ai) Find the circumferences of C1, C2 and C3.
ii) Find the perimeters of triangle P1Q1R1, triangle P2Q2R2
and triangle P3Q3R3

bi) Find the total circumference of all the circles C1, C2, C3, ... .
ii) Find the total perimeter of all the equilateral triangles P1Q1R1, P2Q2R2, P3Q3R3, ... .

Answer :
ai) C1: 16pie cm, C2: 8pie cm, C3: 4pie cm

ii) triangle P1Q1R1: 24 ( surd 3) cm, triangle P2Q2R2: 12( surd 3) cm
Triangle P3Q3R3: 6(surd3) cm

bi) 32pie cm
ii) 48(surd3) cm
更新1:

Give me steps please.

更新2:

Give me steps plz

回答 (1)

2010-11-30 4:38 am
✔ 最佳答案
(a)(i) Let the radius of C1 is x cm. Then since the length of the side of the equilateral triangle P1Q1R1 is √3x cm, the radius of C2 should be x/2 cm by pythagorean theorem. So, the circumferences of C1, C2 and C3 are 16π,8π,4π cm respectively.

(ii) Similarly, the perimeters of equilateral triangles P1Q1R1, P2Q2R2, P3Q3R3 are 3√3x, 3√3x/2, 3√3x/4 cm (with x=8). That is 24√3, 12√3, 6√3 cm

(b)(i) the total circumference of all the circles C1, C2, C3, ... is
a/(1-r)=16π/(1-1/2)=32π cm

(ii) The total perimeter of all the equilateral triangles P1Q1R1, P2Q2R2, P3Q3R3 is a/(1-r)=2(3√3x)=48√3 cm


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