# 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.

Result

v =  7 m

#### Solution:

$a^2 = 3^2 +x^2 \ \\ a^2 = 5^2 + y^2 \ \\ a^2 = 8^2 + (y-x)^2 \ \\ \ \\ 3^2 +x^2 = 5^2 + y^2 \ \\ 3^2 +x^2 = 8^2 + (y-x)^2 \ \\ \ \\ 3^2 +x^2 = 8^2 + y^2-2xy +x^2 \ \\ 3^2 = 8^2 + y^2-2xy \ \\ \ \\ x = (8^2 - 3^2 + y^2) / (2y) \ \\ \ \\ y = 11 / \sqrt{ 3 } \doteq 6.3509 \ m \ \\ \ \\ x = (8^2 - 3^2 + y^2) / (2y) = (8^2 - 3^2 + 6.3509^2) / (2 \cdot \ 6.3509) \doteq 7.5056 \ m \ \\ \ \\ a = \sqrt{ 3^2 + x^2 } = \sqrt{ 3^2 + 7.5056^2 } \doteq 8.0829 \ m \ \\ \ \\ v^2 = a^2 - (a/2)^2 \ \\ \ \\ v = \sqrt{ a^2 - (a/2)^2 } = \sqrt{ 8.0829^2 - (8.0829/2)^2 } = 7 = 7 \ \text { m }$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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Looking for help with calculating roots of a quadratic equation? Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Do you want to convert length units? Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.

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