3.1. The parabola $y=x^{2}$ intersects with the line $y=25$. A circle is constructed on the segment between the intersection points of the parabola and the line as its diameter. Find the area of the convex polygon whose vertices are the intersection points of the given circle and the parabola. Provide the nearest integer to the value of this area.
Answer: 10. Solution. The parabola $y=x^{2}$ intersects the line $y=a$ at points with coordinates $( \pm \sqrt{a} ; a)$. Each of the intersection points of the circle and the parabola mentioned in the problem has coordinates $\left(x ; x^{2}\right)$. For such a point, the distance to the line $y=a$ is $\left|a-x^{2}\right|$, the distance to the y-axis is $|x|$, and the distance to the intersection point of these lines (the center of the circle) is $\sqrt{a}$. By the Pythagorean theorem, $x^{2}+\left(a-x^{2}\right)^{2}=a$, which simplifies to $\left(a-x^{2}\right)\left(a-x^{2}-1\right)=0$. Besides the two points with coordinates $( \pm \sqrt{a} ; a)$, the circle and the parabola also intersect at two points $( \pm \sqrt{a-1} ; a-1)$, which lie on the line $y=a-1$.

Thus, the polygon mentioned in the problem is a trapezoid with height 1 and bases $2 \sqrt{a}$ and $2 \sqrt{a-1}$. Its area is $\sqrt{a}+\sqrt{a-1}$. For $a=25$, we get $5+\sqrt{24} \approx 9.898979 \ldots$.