Question

$\Delta ABC$ is an equilateral triangle of side $2\sqrt{3}cm$. P is any point in the interior of $\Delta ABC$. If x, y, z are the distances of P from the sides of the triangle, then x + y + z =

• A. $2+\sqrt{3}$ cm
• B. 5 cm
• C. 3 cm
• D. 4 cm

Answer 1

The correct option is C

3 cm

$Ar\left(\Delta ABC\right)=Ar(\Delta AOB+\Delta AOC+\Delta BOC)\frac{(2\sqrt{3}{)}^2\sqrt{3}}{4}=\frac{1}{2}x.2\sqrt{3}+\frac{1}{2}y.2\sqrt{3}+\frac{1}{2}z.2\sqrt{3}\frac{12\sqrt{3}}{4}=\frac{1}{2}2\sqrt{3}[x+y+z]3\sqrt{3}=\sqrt{3}[x+y+z]3=x+y+z$