October 27th, 2009, 07:42 PM
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Quote:
Originally Posted by ukorov  Produce a straight line where Y lies on AB, Z lies on DC, and YXZ//AD//BC.
CM = 3DM, hence
CM = 0.75*DC = 0.75*AB
hence
AB / CM
= AB / 0.75AB
= 4/3
= AX / CX
= BX / MX
triangle AYX similiar to CZX (AAA), therefore
AY / CZ = AX / CZ = 4/3
Therefore CZ = BY = (3/7)*AB
area of triangle BXC = BC*BY/2
= BC*(3/7*AB)*1/2
= (BC)(AB)(3/14)
area of rectangle ABCD = (BC)(AB)
Hence, the ratio concerned
= 3:14 | i think this AY / CZ = AX / CZ = 4/3 should be AY / CZ = AX / CX = 4/3
And i dont understand this step Therefore CZ = BY = (3/7)*AB By the way, this qn is only 1 mark.perhaps there a shorter method?
thanks |