KM || FE, MO - ?

Since KM || FE, the trapezoid KMFЕ is isosceles. Therefore, KF = ME = 10. Let R be the radius of the circle. In triangle KFE, by the law of cosines, $$KF^2 = KE^2 + FE^2 - 2 imes KE imes FE imes ext{cos}( ext{angle} KEF)$$. Since KMFЕ is an isosceles trapezoid, the diagonals are equal, so $$KM = FE = 14$$ is incorrect. The diagonals are equal, so $$KE = MF = 10$$. Since KM || FE, the arcs between parallel chords are equal, so arc KF = arc ME. This implies that KF = ME = 10. Let's consider triangle KFE. We have KF = 10, FE = 14. Let's drop a perpendicular from K to FE, let's call the foot P. Then FP = (FE - KM)/2 = (14-2)/2 = 6. In right triangle KPF, $$KP^2 = KF^2 - FP^2 = 10^2 - 6^2 = 100 - 36 = 64$$. So KP = 8. The height of the trapezoid is 8. The radius R of the circumscribed circle can be found using the formula $$R = rac{abc}{4K}$$, where a, b, c are side lengths and K is the area. For triangle KFE, we can use Ptolemy's theorem for cyclic quadrilaterals: $$KM imes FE + KF imes ME = KE imes MF$$. $$2 imes 14 + 10 imes 10 = KE imes KE$$. $$28 + 100 = KE^2$$. $$KE^2 = 128$$. $$KE = ext{sqrt}(128) = 8 ext{sqrt}(2)$$. This is incorrect as KE is given as 10. Let's re-examine the problem. KM || FE. This means the trapezoid KMFЕ is isosceles. So KF = ME = 10. KM = 2, FE = 14. Let's drop perpendiculars from K and M to FE. Let these feet be P and Q respectively. Then FP = QE = (FE - KM)/2 = (14 - 2)/2 = 6. In right triangle KPF, $$KP^2 = KF^2 - FP^2 = 10^2 - 6^2 = 100 - 36 = 64$$. So KP = 8. The height of the trapezoid is 8. The radius R of the circumscribed circle can be found using the formula $$R = rac{a}{2 ext{sin}(A)}$$. In triangle KFE, we can find the angle KEF. Let's use the property that the distance from the center O to a chord is given by $$d = ext{sqrt}(R^2 - (c/2)^2)$$, where c is the chord length. Let $$h_1$$ be the distance from O to KM and $$h_2$$ be the distance from O to FE. Since KM || FE, the distance between the chords is the height of the trapezoid, which is 8. So $$|h_1 ext{ - } h_2| = 8$$. We have $$R^2 = (KM/2)^2 + h_1^2 = (2/2)^2 + h_1^2 = 1 + h_1^2$$. Also, $$R^2 = (FE/2)^2 + h_2^2 = (14/2)^2 + h_2^2 = 49 + h_2^2$$. So $$1 + h_1^2 = 49 + h_2^2$$. $$h_1^2 - h_2^2 = 48$$. $$(h_1 - h_2)(h_1 + h_2) = 48$$. Since the center O is between the chords (as seen in the diagram), $$h_1 + h_2 = 8$$. Substituting this into the equation: $$8(h_1 - h_2) = 48$$, so $$h_1 - h_2 = 6$$. We have a system of equations: $$h_1 + h_2 = 8$$ and $$h_1 - h_2 = 6$$. Adding the two equations gives $$2h_1 = 14$$, so $$h_1 = 7$$. Subtracting the second from the first gives $$2h_2 = 2$$, so $$h_2 = 1$$. We need to find MO. MO is the distance from the center O to the chord KM. So MO = $$h_1$$. Therefore, MO = 7.