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 answer:

v =  7 m

Step-by-step explanation:

$$\begin{gathered} a^2=3^2+x^2 \\ {a}^2=5^2+y^2 \\ {a}^2=8^2+(y-x{)}^2 \\ {3}^2+x^2=5^2+y^2 \\ {3}^2+x^2=8^2+(y-x{)}^2 \\ {3}^2+x^2=8^2+y^2-2x_y+x^2 \\ {3}^2=8^2+y^2-2x_y \\ x=(8^2-3^2+y^2)\mathrm{/}(2y) \\ y=11\mathrm{/}\sqrt{3}\doteq 6.3509\text{ m } \\ x=(8^2-3^2+y^2)\mathrm{/}(2y)=(8^2-3^2+6.3509^2)\mathrm{/}(2\cdot 6.3509)\doteq 7.5056\text{ m } \\ a=\sqrt{3^2+x^2}=\sqrt{3^2+7.5056^2}\doteq 8.0829\text{ m } \\ {v}^2=a^2-(a\mathrm{/}2{)}^2 \\ v=\sqrt{a^2-(a\mathrm{/}2{)}^2}=\sqrt{8.0829^2-(8.0829\mathrm{/}2{)}^2}=7\text{ m} \end{gathered}$$

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