Question 1207821
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Technically, the statement of the problem is deficient, because it does not say that the artist uses the full 50 inches of trim.<br>
Obviously the problem can be solved only if we assume that he does.  So....<br>
(a) If the painting is a square...<br>
Let x be the side length of the painting; then the perimeter is 4x, and the perimeter of the painting plus frame is 4(x+3) = 4x+12.<br>
4x+12 = 50
4x = 38
x = 38/4 = 19/2 = 9.5<br>
ANSWERS:
The painting is a square 9.5 inches on a side.
The frame is a square 9.5+2(3) = 15.5 inches on a side.<br>
(b) If the painting is a rectangle with a length twice its width...<br>
Let x be the width; then the length is 2x; the perimeter is 6x, and the perimeter of the frame is 6x+12.<br>
6x+12 = 50
6x = 38
x = 38/6 = 19/3 = 6 1/3<br>
ANSWERS:
The painting is a rectangle 6 1/3 inches wide and 12 2/3 inches long.
The frame is a rectangle 6 1/3 + 2(3) = 12 1/3 inches wide and 12 2/3 + 2(3) = 18 2/3 inches long.<br>