Question

ABC is an equilateral triangle of side length 10 cm. The minimum distance of vertex A from its opposite side is.

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

Answer 1

The correct option is D $5\sqrt{3}$

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 = 5cm

Now, using pythagorus theorem, in triangle ABD:
$A{D}^2+B{D}^2=A{B}^2$
$A{D}^2+5^2={10}^2$
$A{D}^2+25=100$
$A{D}^2=100-25$
$A{D}^2=75$
$AD=\sqrt{75}$
$AD=5\sqrt{3}$