Question

in an equilateral triangle with area 64√3unit, all the 3altitude are drawn. what is the distance of any of the vertex from the point of intersection of all the 3altitude of the triangle??​

Answer 1

$\bf{\underline{\underline{\bigstar\bigstar \: Solution : }}}$

$\:\:$

$\footnotesize{Area of equilateral triangle = \frac{\sqrt{3}}{4} \(side)^2}$

$\footnotesize{\implies 64\sqrt{3} = \frac{1.72}{4} \(side)^2}$

$\footnotesize{\implies 64 \times 1.72 = 0.43 \times \(side)^2}$

$\footnotesize{\implies \frac{110.08}{0.43} = \(side)^2}$

$\footnotesize{\implies 256 = \(side)^2}$

$\footnotesize{\implies 16 = side}$

$\:\:$

$\footnotesize{height of equilateral triangle = \frac{area}{side}}$

$\footnotesize{\implies height of equilateral triangle = \frac{64\sqrt{3}}{16}}$

$\footnotesize{\implies height of equilateral triangle = \frac{110.08}{16}}$

$\footnotesize{\implies height of equilateral triangle = 6.88}$

$\:\:$

$\footnotesize{required distance = \frac{2}{3} \times 6.88}$

$\footnotesize{required distance = 2 × 2.29}$

$\footnotesize{required distance = 4.58}$

$\bf{\footnotesize{\therefore required distance = 4.58}}$