F.2 - F.3 Maths 一問

2008-07-06 12:25 am

(1) [No figure]
Given 2 circles of areas πr12 and πr22 (r1>r2) respectively. Express, in terms of r1 and r2,
(a) the differece of the square,
(b) the square of the difference,
(c) the sum of the square
of their circumferences. And write your answers in factor form.
(我唔明咩叫 difference of the square 同 square of difference...)
(2)Which is the lower rate: 630 l/h or 150 c.c./s?
(l唔係一，係公升Litre)
(我唔知1升等於等多c.c.)
(3) (A+2)x + B≡ 0 , find the values of A and B.
(4) [Figures aren't drawn to scale]

圖片參考：http://www.au18.net/maths/p23.JPG

In the figure, AB = AC and AD = CD, ∠A=36, CD bisects ∠ACB. Prove that AD=BC.
(5)
圖片參考：http://www.au18.net/maths/P89.JPG

In the figure, △ABC is isosceles, ∠BAD = ∠CAD = ∠ACD. Find ∠ACD.

回答 (1)

用戶10012

2008-07-06 1:55 am

✔ 最佳答案

(1) Let the 2 radius be r and R(R is the bigger one). So their circumferences are 2pr and 2pR respectively (p stands for pi).
(a) Difference of the square of their circumference = (2pR)^2 - (2pr)^2 = 4p^2(R^2 -r^2)= 4p^2(R+r)(R-r).
(b)Square of their difference = (2pR-2pr)^2 = 4p^2(R-r)^2.
(c)Sum of their square= (2pR)^2 + (2pr)^2 = 4p^2(R^2 + r^2).
(2) One litre = 1000 cc. Therefore,
630l/h = 630 x 1000/3600 = 175 cc/s, so 150 cc/s is lower.
(3)(A+2)x +B is identical to 0. That is
A+2 = 0 , so A= -2. And B = 0.

2008-07-05 18:05:25 補充：
(4) Angle A = 36, therefore angle B = (180-36)/2 = 72. AD=DC, therefore, angle ACD = 36, so angle CDB = 72. Therefore, CD=CB, but CD=AD, therefore CB=AD.

2008-07-05 18:10:25 補充：
(5) Let angle BAD =angle CAD = angle ACD = x. Angle B=(180=x)/2. Therefore,
x + x + x + (180-x)/2 = 180. 6x + 180 - x = 360, 5x = 180, therefore, x = 36= angle ACD.

2008-07-05 18:11:19 補充：
Correction: Should be angle B=(180-x)/2.

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