Three parallels

The vertices of an equilateral triangle lie on 3 different parallel lines. The middle line is 5 m and 3 m distant from the end lines. Calculate the height of this triangle.

Correct result:

v =  7 m

Solution:

a2=32+x2 a2=52+y2 a2=82+(y−x)2  32+x2=52+y2 32+x2=82+(y−x)2  32+x2=82+y2−2xy+x2 32=82+y2−2xy  x=(82−32+y2)/(2y)  y=11/3≐6.3509 m  x=(82−32+y2)/(2y)=(82−32+6.35092)/(2⋅ 6.3509)≐7.5056 m  a=32+x2=32+7.50562≐8.0829 m  v2=a2−(a/2)2  v=a2−(a/2)2=8.08292−(8.0829/2)2=7 ma^2=3^2 +x^2 \ \\ a^2=5^2 + y^2 \ \\ a^2=8^2 + (y-x)^2 \ \\ \ \\ 3^2 +x^2=5^2 + y^2 \ \\ 3^2 +x^2=8^2 + (y-x)^2 \ \\ \ \\ 3^2 +x^2=8^2 + y^2-2xy +x^2 \ \\ 3^2=8^2 + y^2-2xy \ \\ \ \\ x=(8^2 - 3^2 + y^2) / (2y) \ \\ \ \\ y=11 / \sqrt{ 3 } \doteq 6.3509 \ \text{m} \ \\ \ \\ x=(8^2 - 3^2 + y^2) / (2y)=(8^2 - 3^2 + 6.3509^2) / (2 \cdot \ 6.3509) \doteq 7.5056 \ \text{m} \ \\ \ \\ a=\sqrt{ 3^2 + x^2 }=\sqrt{ 3^2 + 7.5056^2 } \doteq 8.0829 \ \text{m} \ \\ \ \\ v^2=a^2 - (a/2)^2 \ \\ \ \\ v=\sqrt{ a^2 - (a/2)^2 }=\sqrt{ 8.0829^2 - (8.0829/2)^2 }=7 \ \text{m}

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