h=\frac{a\sqrt3}{2}
a\sqrt3=2h
a=\frac{2h}{\sqrt3}
Pole \Delta
\frac{a^2\sqrt3}{4}=100 \ |*4
podstawiam wyznaczone a
(\frac{2h}{\sqrt3})^2\cdot \sqrt3{}=400
\frac{4\sqrt3h^2}{3}=400\ |*3
4\sqrt3h^2=1200 \ |:4\sqrt3
h^2=\frac{1200}{4\sqrt3}=\frac{300}{\sqrt3}=\frac{300\sqrt3}{3}
h^2=100\sqrt3
h=\sqrt{100\sqrt3}
h=10\sqrt[4]{3} <-- odpowiedź