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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?

 Dec 18, 2018

Best Answer 

 #1
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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

 Dec 18, 2018
 #1
avatar+7763 
+2
Best Answer

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

MaxWong Dec 18, 2018

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