Refer to the attached image.
Given the rectangle ABCD of length 'l' and height 'h'.
Therefore, CD=AB = 'l' and BC = AD = 'h'
We have to determine the area of triangle AEF.
Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE
Area of triangle ADF = [tex]\frac{1}{2}bh[/tex]
= [tex]\frac{1}{2}(DF \times AD)[/tex]
= [tex]\frac{1}{2}(\frac{l}{2} \times h)[/tex]
[tex]=\frac{lh}{4}[/tex]
Area of triangle ECF = [tex]\frac{1}{2}bh[/tex]
= [tex]\frac{1}{2}(CF \times CE)[/tex]
= [tex]\frac{1}{2}(\frac{l}{2} \times \frac{h}{2})[/tex]
[tex]=\frac{lh}{8}[/tex]
Area of triangle ABE = [tex]\frac{1}{2}bh[/tex]
= [tex]\frac{1}{2}(AB \times BE)[/tex]
= [tex]\frac{1}{2}(l \times \frac{h}{2})[/tex]
[tex]=\frac{lh}{4}[/tex]
Now, area of triangle AEF =
Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE
= [tex]72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})[/tex]
= [tex]72 - (\frac{2lh+lh+2lh}{8})[/tex]
=[tex]72 - (\frac{5lh}{8})[/tex]
=[tex]72 - (\frac{5 \times 72}{8})[/tex]
[tex]=\frac{72 \times 8 - (5 \times 72)}{8}[/tex]
= 27 units
Therefore, the area of triangle AEF is 27 units.
