maths mc question

1) ( 4n+1 + 22n ) / 5
= ( 22n+2 + 22n ) / 5
= 22n ( 22 + 1 ) / 5
= 22n [ C ]
2) Let BC = a and the height of the // gram be b.
Area of tri. MDC = ( 0.5a )( b ) / 2 = 6
ab = 24
Area of tri. ABN = ( 0.25a )( b ) / 2 = 0.125 x 24 = 3
Hence area of the shaded region: 24 - 3 - 6 = 15cm2 [ C ]
3) Area of tri. ACN : Area of tri. MCN = 4 : 3
Suppose they are 4A and 3A respectively.
Area of tri. ABM : Area of tri. AMC = 1 : 2
Area of tri. AMC = 3A + 4A = 7A
Hence area of tri. ABM = 7A / 2 = 3.5A
Lastly, area of tri. ABC = 3.5A + 7A = 10.5A
which gives required ratio as 21 : 8 [ A ].

2009-02-28 11:46:11 補充：
For Q2 & 3, the key is to locate pairs of triangles which have same height.
If 2 triangles have same height and their bases are in the ratio of m : n, then the ratio of their areas is m : n.

2009-02-28 12:27:58 補充：
I've sent a letter to you concerning Q16, please check. For the other 2 Qs, please wait, I'll post the solutions later.

2009-02-28 21:56:45 補充：
Please visit http://i238.photobucket.com/albums/ff245/chocolate328154/44480711.jpg for the other 2 Qs.

2009-03-01 11:15:36 補充：
Okay, so you try to calculate as cyclic. quad. I've done that once and the answer should be 2.