Question
Find the altitudes of triangle $\left(-\dfrac{ 3 }{ 2 },~0\right)$ $\left(\dfrac{ 3 }{ 2 },~0\right)$ $\left(0,~2.598\right)$.
Answer
The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = 2.5981 ~~,~~ h_b = 2.5981 ~~,~~ h_c = 2.598 $$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 = 3.897 $ and $ d(B,C) = 2.9999 $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 3.897}{ 2.9999 } = 2.5981 $$This page was created using
Triangle Calculator