7. Triangles $A B C$ and $A B D$ are inscribed in a semicircle with diameter $A B=5$. The perpendicular from $D$ to $A B$ intersects segment $A C$ at point $Q$, ray $B C$ at point $R$, and segment $A B$ at point $P$. It is known that $P R=27 / 10$, and $P Q=5 / 6$. Find the length of segment $D P$. If necessary, round the answer to the hundredths.

![](https://cdn.mathpix.com/cropped/2024_05_06_05e9aaee929c8f4a3d1cg-2.jpg?height=551&width=648&top_left_y=2009&top_left_x=704)
Answer: 1.5.

Solution: Since $AB$ is the diameter, then $\angle ADB = \angle ACD = 90^{\circ}$. Angles $\angle QAB$ and $\angle PRB$ are equal, as they are complemented by angle $CBA$ to $90^{\circ}$. Therefore, triangles $\triangle APQ$ and $\triangle RPB$ are similar by two angles. Thus, $\frac{AP}{PQ} = \frac{RP}{PB}$, from which $AP \cdot PB = PQ \cdot PR = \frac{27}{10} \cdot \frac{5}{6} = \frac{9}{4}$. It remains to recall the property of the height of a right triangle: $DP^2 = AP \cdot PB$, from which $DP = \frac{3}{2}$.