Math
posted by Lily
the right triangle ABC, the altitude from vertex C divides the hypotenuse into two segments, one of length 2 and the other of length 16. Find the perimeter of triangle ABC.

Reiny
C must be the right angle.
You have 3 similar triangle, the original and 2 smaller ones
let the altitude be h
2/h = h/16
h^2 = 32
h = √32
in the smaller rightangled triangle
h^2 + 2^2 = BC^2
BC^2 = 32 + 4 = 36
BC = 6
In the other rightangled triangle:
AC^2 = h^2 + 16^2
= 32 + 256 = 288
AC = √288 = 12√2
perimeter of ABC = 6 + 12√2 + 18
= 24 + 12√2
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