We have a triangle \(\triangle ABC\) and a point \(K\) on segment \(\overline{BC}\) such that \(AK\) is an altitude to \(\triangle ABC\). If \(AK = 6,\) \(BK = 8\), and \(CK = 6,\) then what is the perimeter of the triangle?
First thing we know BC = BK + CK = 14.
Then we use Pythagorean theorem twice to get:
\(AB = \sqrt{AK^2 + BK^2} = 10\)
\(AC = \sqrt{AK^2+CK^2} = 6\sqrt2\)
Therefore,
Perimeter = AB + AC + BC = 14 + 10 + 6√2 = 24 + 6 √ 2
