4.8. The height of a trapezoid, whose diagonals are perpendicular to each other, is 4. Find the area of the trapezoid, given that the length of one of its diagonals is 5.
4.8. Let the diagonal $AC$ of trapezoid $ABCD$ with base $AD$ be equal to 5. Extend triangle $ACB$ to parallelogram $ACBE$. The area of trapezoid $ABCD$ is equal to the area of right triangle $DBE$. Let $BH$ be the height of triangle $DBE$. Then $EH^{2}=BE^{2}-BH^{2}=5^{2}-4^{2}=3^{2}$ and $ED=BE^{2} / EH=25 / 3$. Therefore, $S_{DBE}=ED \cdot BH / 2=50 / 3$.