Najpierw poznaj wzory na:
[tex]\text{P}=\frac{\text{a}^2\sqrt3}{4}[/tex]
[tex]\text{h}=\frac{\text{a}\sqrt3}{2}[/tex]
Pole jest równe 6√3. Nie pozostaje nam nic innego jak:
[tex]\frac{\text{a}^2\sqrt3}{4}=6\sqrt3 \ \ |\cdot4\\\\\text{a}^2\sqrt3=24\sqrt3 \ \ |:\sqrt3\\\\\text{a}^2=24\\\\\text{a}=\sqrt{24}=\sqrt{4\cdot6}=2\sqrt6\\\\\text{h}=\frac{\not2\sqrt6\cdot\sqrt3}{\not2}=\sqrt{6\cdot3}=\sqrt{2\cdot3\cdot3}=\sqrt{2\cdot3^2}=\boxed{3\sqrt2} \ \ (\text{Odp}. \ \text{D})[/tex]
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Najpierw poznaj wzory na:
[tex]\text{P}=\frac{\text{a}^2\sqrt3}{4}[/tex]
[tex]\text{h}=\frac{\text{a}\sqrt3}{2}[/tex]
Pole jest równe 6√3. Nie pozostaje nam nic innego jak:
[tex]\frac{\text{a}^2\sqrt3}{4}=6\sqrt3 \ \ |\cdot4\\\\\text{a}^2\sqrt3=24\sqrt3 \ \ |:\sqrt3\\\\\text{a}^2=24\\\\\text{a}=\sqrt{24}=\sqrt{4\cdot6}=2\sqrt6\\\\\text{h}=\frac{\not2\sqrt6\cdot\sqrt3}{\not2}=\sqrt{6\cdot3}=\sqrt{2\cdot3\cdot3}=\sqrt{2\cdot3^2}=\boxed{3\sqrt2} \ \ (\text{Odp}. \ \text{D})[/tex]