Oplossing week 8

$ \eqalign{\begin{array}{l} \left\{ \begin{array}{l} AD = 20\sqrt 2 \\ BC = 10\sqrt 3 \\ \end{array} \right. \\ \left\{ \begin{array}{l} \frac{{20\sqrt 2 }}{{10}} = \frac{h}{{10 - x}} \Rightarrow x = 10 - \frac{1}{4}\sqrt 2 \cdot h \\ \frac{{10\sqrt 3 }}{{10}} = \frac{h}{x} \Rightarrow x = \frac{1}{3}\sqrt 3 \cdot h \\ \end{array} \right. \\ \frac{1}{3}\sqrt 3 \cdot h = 10 - \frac{1}{4}\sqrt 2 \cdot h \\ h = 16\sqrt 3 - 12\sqrt 2 \\ \end{array}} $

De hoogte is $16\sqrt{3}-12\sqrt{2}$