Arm

Calculate the length of the arm r of isosceles triangle ABC, with base |AB| = 14 cm and a height v=18 cm.

Correct answer:

r =  19.31 cm

Step-by-step explanation:

$$\begin{gathered} a=14\text{ cm } \\ v=18\text{ cm } \\ {r}^2=v^2+{{\displaystyle \frac{a}{2}}}^2 \\ r=\sqrt{(a\mathrm{/}2{)}^2+v^2}=\sqrt{(14\mathrm{/}2{)}^2+18^2}=\sqrt{373}=19.31\text{ cm} \end{gathered}$$

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