In a right-angled triangle, the length of the medians from the vertices of acute angles are 7 cm and $$4\sqrt 6 $$  cm. What is the length of the hypotenuse of the triangle (in cm)?

$$\eqalign{ & AD = 7{\text{ cm}} \cr & CE = 4\sqrt 6 {\text{ cm}} \cr & {\text{Since, }}4\left( {A{D^2} + C{E^2}} \right) = 5A{C^2} \cr & 4\left( {{{\left( 7 \right)}^2} + {{\left( {4\sqrt 6 } \right)}^2}} \right) = 5A{C^2} \cr & 4\left( {49 + 96} \right) = 5A{C^2} \cr & 4\left( {145} \right) = 5A{C^2} \cr & A{C^2} = \frac{{4 \times 145}}{5} \cr & AC = \sqrt {4 \times 29} \cr & AC = 2\sqrt {29} {\text{ cm}} \cr} $$