7. The midline $E F$ of trapezoid $A B C D$ is 15 cm long (see figure), $\angle A B C=\angle A E F=90^{\circ}, G$ is a point on $E F$. If the area of triangle $A B G$ is $1 / 5$ of the area of trapezoid $A B C D$, what is the length of $E G$ in centimeters?
6 cm
7.【Solution】The area of trapezoid $\mathrm{ABCD}$ is equal to $\mathrm{EF} \times \mathrm{AB}$, and the area of triangle $\mathrm{ABG}$ is equal to $\frac{1}{2}$ $\mathrm{EG} \times \mathrm{AB}$. Therefore, the ratio of the area of triangle $\mathrm{ABG}$ to the area of trapezoid $\mathrm{ABCD}$ is equal to the ratio of $\frac{1}{2} \mathrm{EG}$ to $\mathrm{EF}$. According to the conditions given in the problem, the area of triangle $\mathrm{ABG}$ is $\frac{1}{5}$ of the area of trapezoid $\mathrm{ABCD}$, which means $\mathrm{EG}$ is $\frac{2}{5}$ of $\mathrm{EF}$. Since $\mathrm{EF}$ is 15 cm long, the length of $\mathrm{EG}$ is: $15 \times \frac{2}{5}=6$ (cm).

Answer: $\mathrm{EG}$ is 6 cm long.