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

Dear Student,

We form our diagram from given information , As : Here we join PA , PB and PC and height of $∆$ APB = PD = x cm , height of $∆$ BPC = PE = y cm and height of $∆$ APC = PF = z  cm

As given sides of equilateral triangle ABC =  2$\sqrt{3}$ cm , So AB  =  BC  =  AC =  2$\sqrt{3}$ cm

And we know area of equilateral triangle = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^{2}$ , So

Area of $∆$ ABC  =

And we know area of triangle = , So

Area of $∆$ APB = = $\sqrt{3}$ x cm2  ,

Area of $∆$ BPC = = $\sqrt{3}$ y cm2  ,

And

Area of $∆$ APC = = $\sqrt{3}$ z cm2  ,

And

Area of $∆$ APB + Area of $∆$ BPC + Area of $∆$ APC = Area of $∆$ ABC  , Substitute values we get :

$\sqrt{3}$ x + $\sqrt{3}$ y  + $\sqrt{3}$ z  =  3 $\sqrt{3}$  ,

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

x  + y  + z   = 3  cm                                                        ( Ans )