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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?

Guest Dec 18, 2018

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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

MaxWong Dec 18, 2018

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MaxWong · Answer 1 · 2018-12-18T02:16:48

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

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