The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = \frac{\sqrt{ 5 }}{ 5 } ~~,~~ h_b = \frac{ 10 \sqrt{ 157}}{ 157 } ~~,~~ h_c = \frac{ 10 \sqrt{ 97}}{ 97 } $$The altitude $ h_a $ can be found using formula:
$$ A = \frac{ a \cdot h_a }{2} $$Where $ A $ is a triange area and $ a $ is a length of a side $ BC $. In this eample we have: $ A = 5 $ and $ d(B,C) = 10 \sqrt{ 5 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 5}{ 10 \sqrt{ 5 } } = \frac{\sqrt{ 5 }}{ 5 } $$