4. Given an isosceles trapezoid \(ABCD (AD \parallel BC, AD > BC)\). A circle \(\Omega\) is inscribed in angle \(BAD\), touches segment \(BC\) at point \(C\), and intersects \(CD\) again at point \(E\) such that \(CE = 9, ED = 7\). Find the radius of the circle \(\Omega\) and the area of trapezoid \(ABCD\).
Answer: $R=6, S_{A B C D}=96+24 \sqrt{7}$.

Solution. Let the points of tangency of the circle with the sides $A B$ and $A D$ of the trapezoid be denoted as $K$ and $W$ respectively. By the tangent-secant theorem, $D W^{2}=D E \cdot D C=7 \cdot 16, D W=4 \sqrt{7}$. Since $C$ and $W$ are points of tangency of the circle with parallel lines $B C$ and $A D$, the segment $C W$ is the diameter of the circle, perpendicular to these lines. By the Pythagorean theorem from the right triangle $C D W$, we find that $C W=\sqrt{16^{2}-(4 \sqrt{7})^{2}}=12$. Therefore, the radius $R$ of the circle is $\frac{1}{2} C W=6$. Let $B C=x$. Then $B K=x$ (tangents drawn from a point to a circle are equal), and since the trapezoid is isosceles, $A K=A B-B K=16-x$. Thus, $A W=A K=16-x$. From this, we get that the sum of the bases is $B C+A D=x+(4 \sqrt{7}+16-x)=16+4 \sqrt{7}$, and the area of the trapezoid is $\frac{1}{2} \cdot 12 \cdot(16+4 \sqrt{7})=96+24 \sqrt{7}$.