oblicz pole trojkata rownobocznego ktorego
a) bok ma dlugosc a=4 pierwiastek 3
b) wysokosc ma dlugosc h=9dm
dam naj!!​



Odpowiedź :

PUNKT A

[tex]a = 4\sqrt{3}[/tex]

[tex]P = \frac{a^{2}\sqrt{3} }{4} = \frac{(4\sqrt{3})^{2} \sqrt{3} }{4} = \frac{(4^{2} * (\sqrt{3})^{2}) \sqrt{3} }{4} = \frac{16*3*\sqrt{3} }{4} = \frac{48\sqrt{3} }{4} = 12\sqrt{3}[/tex]

PUNKT B

[tex]h = 9 dm[/tex]

[tex]h = \frac{a\sqrt{3} }{2} /*2\\2h = a\sqrt{3} /:\sqrt{3}\\a = \frac{2h}{\sqrt{3} } = \frac{2h*\sqrt{3} }{\sqrt{3} * \sqrt{3} } = \frac{2*9*\sqrt{3} }{3} = \frac{18\sqrt{3} }{3} = 6\sqrt{3}[/tex]

[tex]P = \frac{a^{2} \sqrt{3} }{4} = \frac{(6\sqrt{3})^{2} * \sqrt{3} }{4} = \frac{(6^{2}*(\sqrt{3} )^{2})*\sqrt{3} }{4} = \frac{36*3*\sqrt{3} }{4} = \frac{108\sqrt{3} }{4} = 27\sqrt{3}[/tex]