5. A trapezoid has an area of $80 \mathrm{~cm}^{2}$ and a height of $8 \mathrm{~cm}$. The midpoint of the midline of the trapezoid is 3cm away from one leg and 4cm away from the other leg. Determine the lengths of the bases of this trapezoid.
Solution. Let $SP=3 \text{ cm}$ and $SQ=4 \text{ cm}$ (see the diagram). From the area of the trapezoid, we get $P=mh$, i.e., $80=8m$, from which we find the midline of the trapezoid to be $m=10 \text{ cm}$. Therefore, $a+b=20 \text{ cm}$, and the right triangles $SMP$ and $SMQ$ have hypotenuses of $5 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_66f786bcf1c2286e0245g-4.jpg?height=435&width=771&top_left_y=1484&top_left_x=518)

From the Pythagorean theorem, it follows that $MP=4 \text{ cm}$ and $NQ=3 \text{ cm}$. The right triangles $SMP$ and $BCF$ are similar because they have equal angles at vertices $B$ and $M$, so $SP:MP = CF:BF$, i.e., $3:4 = 8:x$, hence $x=\frac{32}{3}$. Similarly, from the similarity of the triangles, we get $SQ:QN = DE:AE$, i.e., $4:3 = 8:y$, hence $y=6$. Quadrilateral $CDEF$ is a rectangle, so $EF=CD=b$. Therefore, $20=a+b=x+b+y+b=2b+6+\frac{32}{3}$, from which we get $b=\frac{5}{3} \text{ cm}$ and $a=\frac{55}{3} \text{ cm}$.