SOLUTION: An oil painting is 10 inches longer than it is wide and is bordered on all sides by a 3 inch wide frame. If the area of the frame alone is 402 inches squared, what are the dimensio
Question 1104022: An oil painting is 10 inches longer than it is wide and is bordered on all sides by a 3 inch wide frame. If the area of the frame alone is 402 inches squared, what are the dimensions of the painting? Found 2 solutions by josmiceli, ikleyn:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = the width of the painting = the length of the painting = the area of the painting = the area of painting plus frame
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and = length of the painting
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The dimensions are 25.5 x 35.5
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check:
OK
The problem says
L - W = 10, (1)
(L+2*3)*(W+2*3) - L*W = 402. (2)
Simplify (2). It is equivalent to
(L+6)*(W+6) - LW = 402,
LW + 6L + 6W + 36 - LW = 402,
6L + 6W = 402 - 36 = 366,
6*(L+W) = 366 ====> L + W = = 61.
So, the problem is equivalent to these two equations
L - W = 10,
L + W = 61.
------------------Add the equations (both sides). You will get
2L = 10 + 61 = 71 ====> L = = 35.5.
Answer. The dimensions of the oil painting are 35.5 inches (length) and 25.5 inches (width).
