Geometry – Area Chasing 2

$\tan 15 =\dfrac{h}{x}$ $\implies$ $x=\dfrac{h}{\tan 15}$ $\color{#D61F06}(1)$

$\tan 30=\dfrac{h}{5-x}$ $\implies$ $5-x=\dfrac{h}{\tan 30}$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(1)$ in $\color{#D61F06}(2)$ , we have

$5-\dfrac{h}{\tan 15}=\dfrac{h}{\tan 20}$

$5=\dfrac{h}{\tan 30} + \dfrac{h}{\tan 15}$

$h \approx 0.915$

Therefore, the area of $\triangle DEC$ is

$\dfrac{1}{2}(5)(5-0.915) \approx$ $\boxed{10.2}$