Question
Find the altitudes of triangle $\left(0,~0\right)$ $\left(12,~6\right)$ $\left(18,~0\right)$.
Answer
The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = 9 \sqrt{ 2 } ~~,~~ h_b = 6 ~~,~~ h_c = \dfrac{ 18 \sqrt{ 5}}{ 5 } $$Explanation
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 = 54 $ and $ d(B,C) = 6 \sqrt{ 2 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 54}{ 6 \sqrt{ 2 } } = 9 \sqrt{ 2 } $$This page was created using
Triangle Calculator