# Arm

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

Result

r =  19.31 cm

#### Solution:

$a=14 \ \text{cm} \ \\ v=18 \ \text{cm} \ \\ \ \\ r^2=v^2 + \dfrac{ a }{ 2 } ^2 \ \\ \ \\ r=\sqrt{ (a/2)^{ 2 } + v^{ 2 } }=\sqrt{ (14/2)^{ 2 } + 18^{ 2 } } \doteq \sqrt{ 373 } \doteq 19.3132 \doteq 19.31 \ \text{cm}$

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