Question

If the minimum distance from its opposite side of an equilateral triangle is $4\sqrt{3}$ cm, then determine the side length of the triangle.

Answer 1

Length of the perpendicular drawn from vertex A is the minimum distance of vertex A from its opposite side.


Let's assume the side length of the triangle be ${}^{\prime }{a}^{\prime }$.

The perpendular drawn divides the opposite side in two equal parts. Therefore, the length of side, BD = $\frac{a}{2}$cm

Using Pythgorus theorem in triangle ABD;

$A{D}^2+B{D}^2=A{B}^2$

$(4\sqrt{3}{)}^2+(\frac{a}{2}{)}^2=a^2$

$48+\frac{a^2}{4}=a^2$

$48=a^2-\frac{a^2}{4}$

$48=\frac{3a^2}{4}$

$64=a^2$

$a=8cm$