Question
Find the altitudes of triangle $\left(-4,~10\right)$ $\left(8,~-2\right)$ $\left(0,~0\right)$.
Answer
The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = \dfrac{ 36 \sqrt{ 17}}{ 17 } ~~,~~ h_b = \dfrac{ 36 \sqrt{ 29}}{ 29 } ~~,~~ h_c = 3 \sqrt{ 2 } $$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 = 36 $ and $ d(B,C) = 2 \sqrt{ 17 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 36}{ 2 \sqrt{ 17 } } = \frac{ 36 \sqrt{ 17}}{ 17 } $$This page was created using
Triangle Calculator