Question

In rectangle ABCD, AB = 5 and BC = 3. Points F and G are on CD so that DF = 1 and GC = 2. Lines AF and BG intersect at E. Find the area of triangle AEB?

Answer 1

Area of `DeltaAEB=12.5` units.

Explanation:

The above is described in the figure below, the difference being that we have drawn `FK_|_AB`, intersecting `CD` at `H`.

Let `HF=x` and `EH=y`.

Now as `CD`||`AB`, we have similar triangles `EFH` and `EAK` and hence `(HF)/(AK)=(EH)/(EK)` i.e. `x/(1+x)=y/(3+y)`

or `3x+xy=y+xy` or `3x=y` i.e. `x=y/3`

Similarly in similar triangles `EGH` and `EBK` we have `(HG)/(BK)=(EH)/(EK)` i.e. `(2-x)/(2-x+2)=y/(3+y)` or `(2-x)/(4-x)=y/(3+y)`

or `6-3x+2y-xy=4y-xy`

or `2y+3x=6` and as `x=y/3`

`2y+y=6` or `3y=6` i.e. `y=2` and `x=2/3`

As such `EK=EH+HK=2+3=5`

and area of triangle `AEB=(EKxxAB)/2=(5xx5)/2=12.5` units.