Question

determine area de um triângulo equilátero com
b) altura de 6 m.​

Answer 1

$vamos \: relembrar \: algumas \: formulas \: do \: triangulo \: equilatero \\ altura \: do \: triangulo \: equilatero \ \: = \frac{l \sqrt{3} }{2} \\ area \: do \: triangulo \: equilatero = \frac{l {}^{2} \sqrt{3} }{4} \\ = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \\ \frac{l \sqrt{3} }{2} = 6 \: \: = > \: l \sqrt{3} = 6 \times 2 \\ l \sqrt{3} = 12 \: \: = > \: \: l = \frac{12}{ \sqrt{3} \: }m \\ a = \frac{( \frac{12}{ \sqrt{3} }) {}^{2} \sqrt{3} }{4} \\ a = \frac{( \frac{12 {}^{2} }{ \sqrt{3 {}^{2} } }) {}^{} \sqrt{3} }{4} \: \: = > \: \: a = \frac{ \frac{144}{3} \sqrt{3} }{4} \\ a = \frac{48 \sqrt{3} }{4} \: \: = > \: \: a = 12 \sqrt{3} \: m {}^{2} \\ area \: = \: 12 \sqrt{3} \: m {}^{2}$