4. One of the lateral sides of the trapezoid is perpendicular to the bases and equals $2 R$. A circle is constructed on this side as a diameter, which divides the other lateral side into three segments. The ratio of the lengths of these segments is 12:15:5 (counting from the lower base). Find the area of the trapezoid.

Answer. $S=\frac{7 R^{2}}{\sqrt{15}}$.
Let the given trapezoid be $ABCD$ ($AD=2R$, $AB$ - the upper base, points $P$ and $Q$ - the points of intersection of the circle with the lateral side $BC$, $BP: PQ: QC=5: 15: 12$) and let $BP=5x$. Then $PQ=15x$, $QC=12x$, and by the tangent-secant theorem, we find that $BA=\sqrt{BP \cdot BQ}=10x$, $CD=\sqrt{CQ \cdot CP}=18x$.

Let $\mathrm{BH}$ be the height of the trapezoid. Then $\mathrm{CH}=\mathrm{CD}-\mathrm{AB}=8x$ and by the Pythagorean theorem for triangle $\mathrm{BCH}$, we get $\quad(32x)^{2}=(2R)^{2}+(8x)^{2}, \quad$ from which $x=\frac{R}{4\sqrt{15}}$. Then the area of the trapezoid is $S=\frac{AB+CD}{2} \cdot AD=28 \times R=\frac{7R^{2}}{\sqrt{15}}$.