+12 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 ?
+128474
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
+128474
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
