Question

Find the length of the attitude of an equilateral triangle of side
$\sqrt[3]{3}$

Answer 1

Hey friend!

Here's your answer.

$area \: of \: triangle = \frac{ \sqrt{3} }{4} \times ( 3 \sqrt{3} )^{2} cm \\ \frac{ \sqrt{3} }{4} \times 27 \\ = \frac{27 \sqrt{3} }{4} {cm}^{2} \\ let \: the \: height \: be \: h. \\ area \: of \: triangle = \frac{1}{2} bh \\ = \frac{1}{2} \times 3 \sqrt{3} h \\ = \frac{27 \sqrt{3} }{4} = \frac{3 \sqrt{3}h }{2} \\ h = \frac{ 27 \sqrt{3} }{4} \times \frac{2}{ 3\sqrt{3} } \\ = 4.5cm$

Hope this helps!

Answer 2

$area \: of \: triangle = \frac{ \sqrt{3} }{4} \times ( 3 \sqrt{3} )^{2} cm \\ \frac{ \sqrt{3} }{4} \times 27 \\ = \frac{27 \sqrt{3} }{4} {cm}^{2} \\ let \: the \: height \: be \: h. \\ area \: of \: triangle = \frac{1}{2} bh \\ = \frac{1}{2} \times 3 \sqrt{3} h \\ = \frac{27 \sqrt{3} }{4} = \frac{3 \sqrt{3}h }{2} \\ h = \frac{ 27 \sqrt{3} }{4} \times \frac{2}{ 3\sqrt{3} } \\ = 4.5cm$