$ABC$ is an equilateral triangle such that vertices $B,C$ lie on two parallel lines at a distance of 6 units. $A$ lies between the parallel lines at a distance 4 units from one of them. Find the side length of the equilateral triangle.
+1641 ABC is an equilateral triangle such that vertices B,C lie on two parallel lines at a distance of 6 units. A lies between the parallel lines at a distance of 4 units from one of them. Find the side length of the equilateral triangle.
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The side length of a triangle ABC is ≈ 6.11 units
+129907 Imgaine that we have a rectangle with a height of 6
From C, draw a perpendicular to the base of the rectangle
We have three right triangles
sqrt (S^2 - 2^2) is the width of the rectangle
And sqrt (S^2 - 6^2) + sqrt (S^2 - 4^2) also represents the width
So
sqrt (S^2 - 4) = sqrt ( S^2 - 36) + sqrt ( S^2 - 16) square both sides
S^2 - 4 = S^2 - 36 + S^2 -16 + 2 sqrt [ (S^2 -36) ( S^2 - 16) ]
48 - S^2 = 2sqrt [(S^2 - 36) (S^2 -16) ] square both sides again
S^4 - 96S^2 + 2304 = 4 S^4 - 208S^2 + 2304
3S^4 - 112S^2 = 0
S^2 ( 3S^2 - 112) = 0
The second factor will produce a positive result for S
S^2 = 112/3
S = sqrt (112 / 3) = sqrt (16 * 7) / sqrt (3) = 4sqrt (7/3) ≈ 6.11 (as jugoslav found !!!)
