Question

ABC is an equilateral triangle of side length 20 cm. The minimum distance of vertex A from its opposite side is _____.

• A. $4\sqrt{3}$
• B. $5\sqrt{3}$
• C. $10\sqrt{3}$
• D. $8\sqrt{3}$

Answer 1

The correct option is C $10\sqrt{3}$

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


The perpendular drawn divides the opposite side in two equal parts. Therefore, the length of side, BD = 10 cm

Now, using pythagorus theorem, in triangle ABD:
$A{D}^2+B{D}^2=A{B}^2$
$A{D}^2+{10}^2={20}^2$
$A{D}^2+100=400$
$A{D}^2=400-100$
$A{D}^2=300$
$AD=\sqrt{300}$
$AD=10\sqrt{3}$

$\therefore$ $AD=10\sqrt{3}$ is the minimum distance of vertex A from its opposite side.