Question

Determine a aréa de um triangulo equilátero cuja a altura mede 12√3 cm.

Answer 1

$h = \dfrac{l \sqrt{3} }{2} \\ \\ 12 \sqrt{3}\cdot2 = l \sqrt{3} \\ \\ 24 \sqrt{3} ={l \sqrt{3}$

L = 24 cm

$A = \dfrac{l^2 \sqrt{3} }{4} \\ \\ A=\dfrac{576 \sqrt{3} }{4} \\ \\ \boxed{A=144 \sqrt{3} \ cm^2}$