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Calculate the height of an isosceles triangle with base 37.8 mm long and an arm 23.1 mm long.

Correct result:

v =  13.2816 mm

Solution:

$$\begin{gathered} z=37.8\text{ mm } \\ r=23.1\text{ mm } \\ {r}^2=(z\mathrm{/}2{)}^2+v^2 \\ v=\sqrt{r^2-(z\mathrm{/}2{)}^2}=\sqrt{23.1^2-(37.8\mathrm{/}2{)}^2}=13.2816\text{ mm} \end{gathered}$$

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