Question 189213
Let x = distance of the edge of the picture to the edge of the frame
  35-2x = length of the picture 
  25-2x = width of the picture

{{{(35-2x)*(25-2x)=375}}}
{{{875-120x+4x^2=375}}}

Subtracting 375 from both sides, and re-arranging the terms, we get:

{{{4x^2-120x+500=0}}}

Factoring, we get:

{{{(4x-20)*(x-25)=0}}}

So, 

4x-20=0 and/or  x-25=0
4x=20           x=25
x=5

We take x=5, because if x=25, then it would make the length of the picture 
(35-2x)=35-2*25=35-50=-15 and length cannot be negative.

So the edge of the picture is 5cm away from the edge of the frame.