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,

Please find below the solution to the asked query:

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  = $\frac{\sqrt{3}}{4}\times {\left(2\sqrt{3}\right)}^2 = \frac{\sqrt{3}}{4}\times 12 = 3\sqrt{3} {\mathrm{cm}}^2$

And we know area of triangle = $\frac{1}{2}\times \mathrm{Base} \times \mathrm{Height}$ , So

Area of $∆$ APB = $\frac{1}{2}\times \mathrm{AB} \times \mathrm{PD} = \frac{1}{2}\times 2\sqrt{3} \times x$ = $\sqrt{3}$ x cm²  ,

Area of $∆$ BPC = $\frac{1}{2}\times \mathrm{BC} \times \mathrm{PE} = \frac{1}{2}\times 2\sqrt{3} \times y$ = $\sqrt{3}$ y cm²  ,

And

Area of $∆$ APC = $\frac{1}{2}\times \mathrm{AC} \times \mathrm{PF} = \frac{1}{2}\times 2\sqrt{3} \times z$ = $\sqrt{3}$ 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 )


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