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

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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
----------------------------------------------------------





and
= length of the painting
--------------------------
The dimensions are 25.5 x 35.5
-----------------------------------
check:





OK
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
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).



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