\[Схематический\ рисунок.\]
\[Дано:\]
\[ABCD - квадрат;\]
\[AK\ :KB = 1\ :2;\]
\[DM\ :MC = 3\ :1;\]
\[MK = 13\ см.\]
\[Найти:\]
\[\text{AB.}\]
\[Решение:\]
\[1)\ Дополнительное\ построение:\]
\[KH\bot CD;\]
\[H \in CD.\]
\[2)\ В\ ABCD:\]
\[AB = CD = AD = KH = a;\]
\[DH = AK = \frac{1}{3}a,\ \ \ \ \]
\[CM = \frac{1}{4}a;\]
\[MH = a - \frac{1}{3}a - \frac{1}{4}a = \frac{5a}{12}.\]
\[3)\ В\ \mathrm{\Delta}KHM:\]
\[KH^{2} + MH^{2} = KM^{2}\]
\[a^{2} + \left( \frac{5a}{12} \right)^{2} = 13^{2}\]
\[a^{2} + \frac{25}{144}a^{2} = 169\]
\[\frac{169}{144}a^{2} = 169\]
\[a^{2} = 144\ \ \]
\[a = 12\ см.\]
\[Ответ:\ \ 12\ см.\]