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January 30, 2015

January 30, 2015

Posted by **Lily** on Saturday, December 7, 2013 at 8:40am.

- Math -
**Reiny**, Saturday, December 7, 2013 at 11:11amC 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 right-angled triangle

h^2 + 2^2 = BC^2

BC^2 = 32 + 4 = 36

BC = 6

In the other right-angled 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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