\[Схематический\ рисунок.\]
\[Дано:\ \ \]
\[BD - медиана;\]
\[P_{\text{ABD}} = 32\ см;\]
\[P_{\text{CBD}} = 36\ см;\]
\[BD = 10\ см.\]
\[Найти:\]
\[P_{\text{ABC}}.\]
\[Решение:\]
\[1)\ В\ \mathrm{\Delta}ABD:\]
\[P_{\text{ABD}} = AB + BD + AD\]
\[32 = AB + 10 + AD\]
\[AB + AD = 22.\]
\[2)\ В\ \mathrm{\Delta}CBD:\]
\[P_{\text{CBD}} = CB + BD + CD\]
\[36 = CB + 10 + CD\]
\[CB + CD = 26.\]
\[3)\ В\ \mathrm{\Delta}ABC:\]
\[P_{\text{ABC}} = AB + BC + AC =\]
\[= AB + BC + AD + CD =\]
\[= 22 + 26 = 48\ см.\]
\[Ответ:\ \ 48\ см.\]