Question

in an equilateral triangle ABC, D is any point inside the triangle which is at a distance of 3,4,5 from each vertices .find the area of triangle.

Answer 1

Let [math]\triangle[/math] ABC be the equilateral triangle, and the point Pin it is such that the perpendicular distances from the sides BC, AB and AC are PD=3 cm, PE=4 cm and PF=5 cm respectively.

Let the length of each side of the equilateral triangle be L cm.



Now we have,

[math]\Delta ABC = \Delta PBC + \Delta PAB + \Delta PAC[/math]

[math]= \frac{1}{2} PD.AC + \frac{1}{2} PE.AB + \frac{1}{2} PF.AC[/math]

[math]= \frac{L}{2} (PD+PE+PF)[/math]

[math]= \frac{L}{2}(3 + 4 + 5) = 6L\ cm^2 .[/math]

Now, since [math]\Delta ABC[/math] is an equilateral triangle with side [math]L[/math], its area is given by the formula [math]\frac{\sqrt{3}}{4}L^2[/math].

So [math]\frac{\sqrt{3}}{4}L^2 = 6L[/math].

[math]\implies L = 8\sqrt{3}[/math]

Hence the area of the given equilateral triangle is [math]48[/math][math]\sqrt{3}[/math] [math]cm^2[/math].

Hope it helps