Question

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the area of the inscribed triangle aef.

Answer 1

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 = $\frac{1}{2}bh$

= $\frac{1}{2}(DF \times AD)$

= $\frac{1}{2}(\frac{l}{2} \times h)$

$=\frac{lh}{4}$

Area of triangle ECF = $\frac{1}{2}bh$

= $\frac{1}{2}(CF \times CE)$

= $\frac{1}{2}(\frac{l}{2} \times \frac{h}{2})$

$=\frac{lh}{8}$

Area of triangle ABE = $\frac{1}{2}bh$

= $\frac{1}{2}(AB \times BE)$

= $\frac{1}{2}(l \times \frac{h}{2})$

$=\frac{lh}{4}$

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

= $72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})$

= $72 - (\frac{2lh+lh+2lh}{8})$

=$72 - (\frac{5lh}{8})$

=$72 - (\frac{5 \times 72}{8})$

$=\frac{72 \times 8 - (5 \times 72)}{8}$

= 27 units

Therefore, the area of triangle AEF is 27 units.