$
\eqalign{
& AD = 20\sqrt 2 \,\,\,en\,\,\,BC = 10\sqrt 3 \cr
& {\text{geeft:}} \cr
& \frac{{20\sqrt 2 }}
{{10}} = \frac{h}
{{10 - x}} \Rightarrow x = 10 - \frac{1}
{4}\sqrt 2 \cdot h \cr
& {\text{en}} \cr
& \frac{{10\sqrt 3 }}
{{10}} = \frac{h}
{x} \Rightarrow x = \frac{1}
{3}\sqrt 3 \cdot h \cr
& {\text{dus:}} \cr
& \frac{1}
{3}\sqrt 3 \cdot h = 10 - \frac{1}
{4}\sqrt 2 \cdot h \cr
& h = 16\sqrt 3 - 12\sqrt 2 \cr}
$ |
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