[ Area of a sphere and its parts ]

Segment $E F$ is parallel to the plane in which rectangle $A B C D$ lies, with $E F=3, B C=5$. All sides of rectangle $A B C D$ and segments $A E, B E, C F, D F, E F$ are tangent to a certain sphere. Find the surface area of this sphere.
The section of the sphere by the plane of the rectangle $ABCD$ is a circle inscribed in this rectangle (Fig.1). Therefore, the rectangle $ABCD$ is a square with a side of 5, and the sphere touches its sides at their midpoints. Consider the section of the sphere by the plane $ABE$ (Fig.2). We get a circle inscribed in the triangle $ABE$, and the point $M$ of tangency of this circle with the side $AB$ is the point of tangency of the sphere with this side, i.e., the midpoint of $AB$. If the sphere touches the segments $AE$ and $BE$ at points $P$ and $Q$ respectively, then

$$
AP = AM = \frac{5}{2}, BQ = BM = \frac{5}{2}, EP = EQ
$$

Therefore, $AE = BE$. This means that the triangle $AEB$ is isosceles. Its median $EM$ is perpendicular to the side $AB$. Similarly, the median $FN$ of the triangle $DFC$ is perpendicular to the side $CD$. Let the plane $\alpha$, passing through the intersecting lines $ME$ and $MN$, intersect the plane $DFC$ along the line $NF1$. The line $AB$ is perpendicular to the plane $\alpha$, since $AB \perp MN$ and $AB \perp ME$. The lines $AB$ and $CD$ are parallel, so the line $CD$ is also perpendicular to the plane $\alpha$. Therefore, $CD \perp NF1$, and since $CD \perp NF$ and the lines $NF1$ and $NF$ lie in the plane $DFC$, they coincide. Thus, the line $NF$ also lies in the plane $\alpha$. Therefore, through the line $EF$, parallel to the plane $ABCD$, there is a plane $\alpha$ intersecting the plane $ABCD$ along the line $MN$. Therefore, $EF \| MN$. Since the sphere passes through the midpoints $G$ and $H$ of the sides $AD$ and $BC$ of the square $ABCD$, its center lies in the plane passing through the midpoint of the segment $GH$ perpendicular to the line $GH$, i.e., in the plane $\alpha$. Therefore, the radius $R$ of the sphere is equal to the radius of the circle passing through the vertices $M$ and $N$ of the trapezoid $MEFN$ and touching the smaller base $EF$ of the trapezoid at some point $K$. Let the sphere touch the segment $DF$ at the point $L$, and the extensions of the lateral sides $AE$ and $DF$ of the trapezoid $AEFD$ intersect at the point $T$. Then $TP = TL$ as segments of tangents drawn to the sphere from one point. Moreover, $AP = AG = DG = DL$. Therefore, $AT = DT$, and since $EF \| AD$, then $TE = TF$ and

$$
KE = PE = TP - TE = TL - TF = FL = FK.
$$

Therefore, $K$ is the midpoint of the base $EF$ of the trapezoid $MEFN$. In the isosceles triangle $AEB$ (Fig.2), it is known that

$$
AB = 5, BE = AE = AP + PE = AG + EK = \frac{5}{2} + \frac{3}{2} = 4,
$$

$$
ME = \sqrt{AE^2 - AM^2} = \sqrt{16 - \frac{25}{4}} = \frac{\sqrt{39}}{2}.
$$

Similarly, we find that $FN = \frac{\sqrt{39}}{2}$. Let $EE1$ and $KK1$ be the heights of the isosceles trapezoid $MEFN$ (Fig.3). Then

$$
ME1 = \frac{1}{2}(MN - EF) = \frac{1}{2}(5 - 3) = 1,
$$

$$
KK1 = EE1 = \sqrt{ME^2 - ME1^2} = \sqrt{\frac{39}{4} - 1} = \frac{\sqrt{35}}{2},
$$

$$
MK = \sqrt{KK1^2 + MK1^2} = \sqrt{\frac{35}{4} + \frac{25}{4}} = \sqrt{15},
$$

$$
\sin \angle KMK1 = \frac{KK1}{MK} = \frac{\sqrt{35}/2}{\sqrt{15}} = \frac{\sqrt{21}}{6}.
$$

Since $R$ is the radius of the circle circumscribed around the isosceles triangle $MKN$, then

$$
R = \frac{KN}{2 \sin \angle KMN} = \frac{MK}{2 \sin \angle KMK1} = \frac{\sqrt{15}}{2 \cdot \frac{\sqrt{21}}{6}} = \frac{3 \sqrt{35}}{7}.
$$

Therefore, the surface area of the sphere is

$$
4 \pi R^2 = 4 \pi \left(\frac{3 \sqrt{35}}{7}\right)^2 = \frac{180 \pi}{7}
$$

## Answer

$180 \pi$