MP || NK, MN = PK, MP - NK = 6, S_MPKN = ?

The figure MPKN is a trapezoid since MP || NK. Given MN = PK, it is an isosceles trapezoid. The lengths of the tangent segments from a vertex to the inscribed circle are equal. Let the points of tangency be A on MP, B on NK, C on MN, and D on PK. Then MA = MB, NA = NB, PC = PD, KC = KD. From the image, we can infer that the segments from M to the circle are of length 5, and from P to the circle are of length 5. Thus, MP = MA + AP = 5 + 5 = 10. Since MP || NK and the trapezoid is isosceles, the non-parallel sides MN and PK are equal. Also, the segments from N and K to the circle are equal. Let these lengths be x. Then MN = NC + CK = x + x = 2x and PK = PD + DK = 5 + x. Since MN = PK, we have 2x = 5 + x, which implies x = 5. Therefore, MN = PK = 2 * 5 = 10. The problem states MP - NK = 6. We found MP = 10. So, 10 - NK = 6, which means NK = 4. The height of the trapezoid can be found by considering a right triangle formed by drawing a perpendicular from N to MP. Let this height be h. The difference between the parallel sides is MP - NK = 10 - 4 = 6. This difference is distributed equally on both sides of the isosceles trapezoid. So, the base of the right triangle is (10 - 4) / 2 = 3. The hypotenuse is MN = 10. Using the Pythagorean theorem, h^2 + 3^2 = 10^2, so h^2 + 9 = 100, h^2 = 91, and h = sqrt(91). The area of the trapezoid is S_MPKN = (MP + NK) * h / 2 = (10 + 4) * sqrt(91) / 2 = 14 * sqrt(91) / 2 = 7 * sqrt(91).