2. Given a rectangle $A B C D$ with side lengths $|A B|=16 \mathrm{~cm}$ and $|B C|=12 \mathrm{~cm}$. The diagonals of the rectangle intersect at point $S$. On the side $\overline{A B}$ of rectangle $A B C D$, there is a point $E$ which is equally distant from vertex $A$ and point $S$. Calculate the area of triangle $\triangle A E S$.
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_f1ebd08c743b4a28ce27g-01.jpg?height=559&width=694&top_left_y=1905&top_left_x=290)

Sketch (with point $F$, the midpoint of side $\overline{A B}$):

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For the right triangle $\triangle E F S$, the Pythagorean theorem holds:

$|E S|^{2}=|E F|^{2}+|F S|^{2}(*)$

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It is given that $|F S|=\frac{1}{2}|B C|=6 \mathrm{~cm}$.

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If we denote $|A E|=|E S|=x$, then:

$|E F|=|A F|-|A E|=\frac{1}{2}|A B|-|A E|=8-x$.

Substitute this into (*) and solve:

$x^{2}=(8-x)^{2}+6^{2}$

$x^{2}=64-16 x+x^{2}+36$

$16 x=100$

$x=\frac{25}{4}=6.25 \mathrm{~cm}$

Thus, $|A E|=6.25 \mathrm{~cm}$.

In triangle $\triangle A E S$, the length of side $\overline{A E}$ and the length of the height $\overline{F S}$ to that side are known.

The area is $P=\frac{|A E| \cdot|F S|}{2}=\frac{6.25 \cdot 6}{2} \mathrm{~cm}^{2}$.

The area of triangle $\triangle A E S$ is $P=18.75 \mathrm{~cm}^{2}$.

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