Question

Determine o perímetro de um triângulo equilátero cuja altura mede 27

Answer 1

$x^{2} =( \frac{1}{2} ) ^{2}+27 ^{2} x^{2} = \frac{ x^{2} }{4} +729 4 x^{2} = x^{2} +729 4 x^{2} - x^{2} =729 3 x^{2} =729 x^{2} = \frac{729}{3} x^{2} =243 x= \sqrt{243} x= \sqrt{81} * \sqrt{3} x=9.<strong></strong> \sqrt{3} p=3*x p=3*9 \sqrt{3} p=27 \sqrt{3} Perimetro$