MK = NK = 26, MN = 20, OE = ?

The triangle KMN is isosceles since MK = NK. The segment OE is the radius of the inscribed circle, and E is the midpoint of MN. Thus, triangle KOE is a right-angled triangle. ME = MN/2 = 20/2 = 10. Using the Pythagorean theorem in triangle KME, we find the height KE = sqrt(MK^2 - ME^2) = sqrt(26^2 - 10^2) = sqrt(676 - 100) = sqrt(576) = 24. The area of triangle KMN is (1/2) * MN * KE = (1/2) * 20 * 24 = 240. The semi-perimeter s = (MK + NK + MN) / 2 = (26 + 26 + 20) / 2 = 72 / 2 = 36. The radius of the inscribed circle OE = Area / s = 240 / 36 = 20/3.