If the sides of a triangle measure 8,9,and 10,find the longest side of a similar triangle whose perimeter is 81?

Similar triangles bear the same ratio of the correspoding
sides.

If a , b and c are the sides of one triangle and  d, e and f
are he correspoding sides of the similar triangle , then :

a/d =
b/e= c/f  each = k say....(1)

 Then a = dk, b = ek and c =
fk.

Therefore (a+b+c) = (d+e+f)k.

Or
(a+b+c)/(d+e+f) = k.

Given a=8, b=9 and c = 10.  So a+b+c= 8+9+10 =
27. Also d+e+f = 81.

Therefore (a+b+c)/(d+e+f) = 27/81 =
1/3.

 Therefore from (1) :

d = a/k =
8/(1/3) = 24

e = b/k = 9/(1/3) = 27.

f
= c/k = 10/(1/3) = 30.

So the  sides of the similar triangle whose
perimeter is 81 are: 24,27 and 30.