Question 192725
the  isosceles  triangle  with  one  angle  of  150 deg,  tells  us  that  the  vertex  angle  is  150,  side  angles  are  each  15  deg ,  for  a  total  of  180 deg  (150+15+15)

If  side  angle  was  150,  2*150=300,  oversize  for  triangle  (180 deg)

If  we  sketch  this  triangle,  and  include  an  altitude  h,  base  =(2b),  and  sides =a

we  can  figure  Area  =9(given) =  (1/2) base  *  height

9=(1/2) (2b) (h)

but  height  is  related  to  base  thru  150 deg  angle,  or  15 deg  angle.

from  sketch ( just  make  a  rough sketch,  obtuse  isosceles  triangle)  h/b = tan  15
or  h=(tan15)(b)=.268(b),  substituting  into  Area  eqn

9=(1/2) (2b) (.268b)=.268 *b^2

dividing  both  sides  by  .268

9/.268 =b^2
33.582=b^2

taking  sq  rt
5.795  =b

but  h=.268b=.268*5.795=1.5529

last  side  a  can  be  found  from,  h/a =  sin 15,  or  a=h/sin15 =1.5529/.2588=6

sides  of  triangle  are  a,  a,  and  2b,  or  6,6,11.590
per =  sum  of  sides  =  23.590  ANSWER

checking  area  ,  a=(1/2) (11.590) (1.5529) =9  ok

to  check  sides,  pythagorous  can  be  used,  6^2=1.5529^2 + 5.795^2=6^2  ok