14. The isosceles trapezoid $A B C D$, with bases $A B$ and $C D$, is inscribed in a circle of radius $13 \mathrm{~m}$. It is known that the center of the circle is inside the trapezoid $A B C D$ and that $\overline{A B}=24 \mathrm{~m}, \overline{C D}=10 \mathrm{~m}$. What is the area of $A B C D$?
(A) $272 \mathrm{~m}^{2}$
(B) $289 \mathrm{~m}^{2}$
(C) $170 \mathrm{~m}^{2}$
(D) $306 \mathrm{~m}^{2}$
(E) $340 \mathrm{~m}^{2}$
(14) The correct answer is $(B)$.

Let $O$ be the center of the circumcircle of the trapezoid, $H$ the midpoint of $A B$, and $K$ the midpoint of $C D$. Using the Pythagorean theorem, we get (measuring in $\mathrm{m}$) $O H=\sqrt{O B^{2}-B H^{2}}=\sqrt{13^{2}-12^{2}}=5$ and $O K=\sqrt{O C^{2}-C K^{2}}=\sqrt{13^{2}-5^{2}}=12$. The height of the trapezoid is therefore $O H+O K=12+5=17$ and its area is $\frac{(24+10) \cdot 17}{2}=289$.

Problem proposed by Camilla Casamento Tumeo.