Question

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

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

Answer 1

Answer: (C)

$3$cms

Given : $△ABC$ is equilateral triangle of side $2\sqrt{3}$cms.

Let $P$ be any interior point of $△ABC$ such that $x,y,z$ are the distances of $P$ from the sides of a triangle.

Now, $A\left(△ABC\right)=A\left(△APB\right)+A\left(△APC\right)+A\left(△BPC\right)$

$=\frac{1}{2}\times PE\times AB+\frac{1}{2}\times PG\times AC+\frac{1}{2}\times PF\times BC$

$=\frac{1}{2}\times PE\times 2\sqrt{3}+\frac{1}{2}\times PG\times 2\sqrt{3}+\frac{1}{2}\times PF\times 2\sqrt{3}$

$=(x+y+z)\sqrt{3}$

$\therefore \frac{\sqrt{3}}{4}\times (2\sqrt{3}{)}^2=\sqrt{3}(x+y+z)$

$\therefore x+y+z=3$cms