1. **Problem statement:** We have a right triangle PQR with a right angle at point S on the base PR. Given lengths are $PQ=5$ and $PR=7$. We need to find the length $PS$. 2. **Understanding the problem:** Since $QS$ is perpendicular to $PR$, $S$ is the foot of the altitude from $Q$ to $PR$. This creates two right triangles: $PQS$ and $QSR$. 3. **Key formula:** In a right triangle, the altitude to the hypotenuse relates the segments it creates. Specifically, if $PS = x$, then $SR = PR - PS = 7 - x$. 4. **Using the geometric mean theorem:** The altitude $QS$ satisfies: $$QS^2 = PS \times SR = x(7 - x)$$ 5. **Using Pythagoras in triangle $PQS$:** $$PQ^2 = PS^2 + QS^2$$ $$5^2 = x^2 + QS^2$$ $$25 = x^2 + QS^2$$ 6. **Substitute $QS^2$ from step 4:** $$25 = x^2 + x(7 - x) = x^2 + 7x - x^2 = 7x$$ 7. **Simplify and solve for $x$:** $$25 = 7x$$ $$x = \frac{25}{7} \approx 3.57$$ 8. **Answer:** The length $PS$ is approximately $3.57$.