Points E and F lie on the sides AB and AD respectively of rectangle ABCD
(not the corners). The line segments BF, FC, CE, and ED divide ABCD into
six triangles and two quadrilaterals. BF intersects CE and DE at points P
and Q respectively. CF intersects ED at point R.
Let [WX...Z] denote the area of polygon WX...Z.
If [AEQF] = 35, [BPE] = 9, and [DFR] = 6; then what is [CRQP] ?
let
[EPQ]=a
[QRF]=b
[CRD]=c
[BCP]=d
[CPQR]=e
then we want to fine e
now let AD=m, AB=n, AF=s, and AE=t
then we have that
a+b+c+d+e+50=mn
we also have using the areas of each of ABF, AED, CFD, and CBE
a=(ns-88)/2
b=(mt-82)/2
c=(n(m-s)-12)/2
d=(m(n-t)-18)/2
putting this into equation above we get
2mn=ns-88+mt-82+n(m-s)-12+m(n-t)-18+2e+100
expanding and eliminating we end up with
2e=100
thus
e=50
thus
[CRQP]=50
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Posted by Daniel
on 2009-07-25 15:21:25 |