squences急

bi) Let CkDk cut B1E at Xk.
As △AB1E is a right-angled triangle, if CkDk not cut B1E, then DkE > CkDk, ie
AE - ADk > ABk
==> 15 - (1.7 + 0.3k) > 15.6 - 0.6k . . . (from part ai)
==> 0.3k > 2.3
==> k > 7.666
Therefore, the possible values of k are 8, 9, 10, . . ., 25.

bii) The total area of the red region is :
(AD8)(C1D1)+(D8D9)(C9D9)+(D9D10)(C10D10)+...+(D24D25)(C25D25)
= 4.1*10.8 + 0.3*10.2 + 0.3*9.6 + ... + 0.3*0.6
= 44.28 + 0.3*(10.2 + 0.6)*(25 - 9 + 1)/2
= 71.82 (cm^2)

2012-06-02 14:25:28 補充：
AD1, AD2, ... , AD25 are formed with common difference of 0.3cm, so
D1D2 = D2D3 = ... = D24D25 = 0.3

rectangle ABkCkDk is inside triangle AB1E only when
ADk > ABk
2 + (k - 1)(0.3) > 15 + (k - 1)(-0.6)
1.7 + 0.3k > 15.6 - 0.6k
0.9k > 13.9
k > 15.xxx
thus k = 16

其餘我想你懂得怎樣做^^