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 7:21:27 (counting from the upper base). Find the area of the trapezoid.

Answer. $S=\frac{100 R^{2}}{11 \sqrt{21}}$
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 $BC$, $BP: PQ: QC=7: 21: 27$) and let $BP=7x$. Then $PQ=21x$, $QC=27x$, and by the tangent-secant theorem, we find that $BA=\sqrt{BP \cdot BQ}=14x$, $CD=\sqrt{CQ \cdot CP}=36x$.

Let $BH$ be the height of the trapezoid. Then $\mathrm{CH}=\mathrm{CD}-\mathrm{AB}=22x$ and by the Pythagorean theorem for triangle $\mathrm{BCH}$, we get $(55x)^{2}=(2R)^{2}+(22x)^{2}$, from which $x=\frac{2R}{11\sqrt{21}}$. Then the area of the trapezoid is $S=\frac{AB+CD}{2} \cdot AD=50 \times R=\frac{100R^{2}}{11\sqrt{21}}$.