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Point P is inside equilateral triangle ABC such that the altitudes from P to AB, BC , and CA have lengths 5, 6, and 7 respectively. What is the area of triangle ?

MeepMeep  Jul 16, 2017
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#1
+83935
+3

If I understand the problem correctly, there will be three triangles formed, each with an equal base  which will be  a side length of ABC and respective altitudes of 5,6  and 7

So.....the total  area of these three triangles can be represented as :

(1/2) s ( 5 + 6 + 7 )  = (1/2) s ( `18)  =  9s    (1)

Where s is the side length of the equilateral triangle ABC

And the area of an equilateral triangle can be represented as

(1/2) s^2 ( √3 / 2 )  =  (√3 / 4) s^2        (2)

Set (1)  = (2)  and solve for s

9s =  (√3 / 4) s^2      divide by s

9 = (√3 / 4) s     multiply both sides by  4/ √3

9 *  4 / √3  =  s

36 / √3  =  s  =   12√3

So......using (1)...... the area of ABC  is   9 (12√3)  =  108√3  sq units

CPhill  Jul 16, 2017
#2
+83935
+3

BTW....it is easy to prove that, at a minimum, at least one point "P" exists

Let the y coordinate of our  P  = 6

And using the equation for AB, y = √3x,   and the formula for the distance from a point to a line....let the distance from P to AB = 5.....so we have

y = √3x   →  √3x - y  =   0

abs ( Ax + By + C) / √ [ A^2 + B^2 ]       where A = √3, B = -1   and  C = 0

l √3x - 1(6) l /  2  = 5 → [√3x - 1(6)] = 10 → √3x = 16 →  x = 16 / √3

So...given the specified altitudes to the respective sides,  the value for P  =

( 16 / √3,  6 )

Here's the approximate pic with the three triangles APB, BPC and APC....and  EP, FP and GP are the required altitudes to each side

CPhill  Jul 16, 2017
edited by CPhill  Jul 16, 2017
edited by CPhill  Jul 17, 2017
edited by CPhill  Jul 17, 2017

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