Дано:
Найти: Расстояние от точки M до плоскости треугольника ABC.
Решение:
\( p = \frac{AB + BC + AC}{2} = \frac{13 + 13 + 10}{2} = \frac{36}{2} = 18 \) см.
\[ S = \(\sqrt{p(p-a)(p-b)(p-c)}\) \)
где a=13, b=13, c=10.\[ S = \(\sqrt{18(18-13)(18-13)(18-10)}\) = \(\sqrt{18 \cdot 5 \cdot 5 \cdot 8}\) = \(\sqrt{2 \cdot 9 \cdot 25 \cdot 8}\) = \(\sqrt{16 \cdot 9 \cdot 25}\) = 4 \(\cdot\) 3 \(\cdot\) 5 = 60 \) см².
\[ r = \(\frac{S}{p}\) = \(\frac{60}{18}\) = \(\frac{10}{3}\) \) см.
\[ H^2 + r^2 = h^2 \)
\[ H^2 = h^2 - r^2 \)
\[ H^2 = \(\left\)\(\frac{8 \cdot 3 + 2}{3} \right\)^2 - \(\left\)\(\frac{10}{3} \right\)^2 = \(\left\)\(\frac{26}{3} \right\)^2 - \(\left\)\(\frac{10}{3} \right\)^2 \)
\[ H^2 = \(\frac{26^2 - 10^2}{3^2}\) = \(\frac{676 - 100}{9}\) = \(\frac{576}{9}\) \)
\[ H = \(\sqrt\){\(\frac{576}{9}\)} = \(\frac{24}{3}\) = 8 \) см.
Ответ: 8 см