[THOUGHT] Set the cylinder with bottom base in plane z=0 and top base in plane z=h, radius R=5. Let chord AB=5 on the bottom and chord CD=3 on the top. The condition AD=BC forces the midpoints of chords AB and CD to lie on the same diameter (or opposite), hence AB ∥ CD and A,B,C,D are coplanar; ABCD is a trapezoid with bases AB=5 and CD=3.

Let d1 be the distance from the center to chord AB and d2 the distance from the center to chord CD: d1 = sqrt(R^2-(5/2)^2), d2 = sqrt(R^2-(3/2)^2). In the plane of the trapezoid, the distance between the parallel lines containing AB and CD equals sqrt((d2-d1)^2 + h^2). So area(ABCD) = (5+3)/2 * sqrt((d2-d1)^2 + h^2) = 4*sqrt((d2-d1)^2 + h^2).

Let H be the projection of D onto z=0. Then triangle ABH lies in z=0; its base is AB=5 and its altitude to AB equals |d2-d1|, so area(ABH) = (1/2)5|d2-d1|.

Given area(ABCD)=4*area(ABH), solve for h^2 and simplify to k. [/THOUGHT] [PYTHON] import sympy as sp

R = sp.Integer(5) d1 = sp.sqrt(R**2 - (sp.Rational(5,2))2) # distance center->chord AB d2 = sp.sqrt(R2 - (sp.Rational(3,2))**2) # distance center->chord CD delta = sp.simplify(d2 - d1)

h2 = sp.simplify(sp.Rational(21,4) * delta**2) # from sqrt(delta^2+h^2)=(5/2)delta print("d1 =", d1) print("d2 =", d2) print("delta =", delta) print("h^2 simplified =", h2) print("h^2 expanded =", sp.expand(h2)) print("h^2 numeric =", sp.N(h2, 30))

Verify by direct coordinates and area computation in 3D using Newell's method (planar polygon area)

h = sp.Symbol('h', positive=True)

theta = sp.acos(1 - (sp.Rational(5,1)2)/(2*R2)) # not needed; use d1 computed. phi = 2*sp.asin(sp.Rational(3,10))

Coordinates with chord midpoints on +x-axis, chords vertical

xA = d1 A = sp.Matrix([xA, -sp.Rational(5,2), 0]) B = sp.Matrix([xA, sp.Rational(5,2), 0])

xC = d2 C = sp.Matrix([xC, sp.Rational(3,2), h]) D = sp.Matrix([xC, -sp.Rational(3,2), h])

H = sp.Matrix([xC, -sp.Rational(3,2), 0])

def triangle_area(P,Q,Rp): return sp.sqrt(((Q-P).cross(Rp-P)).dot((Q-P).cross(Rp-P)))/2

Newell method for planar polygon area magnitude

def polygon_area(points): nx = ny = nz = 0 for i in range(len(points)): x1,y1,z1 = points[i] x2,y2,z2 = points[(i+1)%len(points)] nx += (y1 - y2)(z1 + z2) ny += (z1 - z2)(x1 + x2) nz += (x1 - x2)*(y1 + y2) return sp.sqrt(nx2 + ny2 + nz**2)/2

area_quad = sp.simplify(polygon_area([A,B,C,D])) area_tri = sp.simplify(triangle_area(A,B,H)) print("Area quad (symbolic) =", area_quad) print("Area tri ABH (symbolic) =", area_tri)

eq = sp.Eq(area_quad, 4*area_tri) sol_h2 = sp.solve(eq, h**2) print("Solve for h^2 from area equation:", sol_h2) print("Check match:", sp.simplify(sol_h2[0] - h2)) [/PYTHON]

[THOUGHT] From the verified Python output, (h^2 = \frac{1743}{8}-\frac{105\sqrt{273}}{8}), so (k=\frac{1743-105\sqrt{273}}{8}). [ \boxed{\frac{1743-105\sqrt{273}}{8}} ]