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}}
$
q9669img1.gif

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

©2004-2019 W.v.Ravenstein