Question 256781: If the length and width of a rectangle were each increased by 1, the area would be
84. The area would be 48 if the length and width were each diminished by 1.
Find the perimeter of the original rectangle.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let L = length and W = width.
area = L*W
(L+1)*(W+1) = 84
(L-1)*(W-1) = 48
multiply these out to get:
LW + L + W + 1 = 84
LW - L - W + 1 = 48
add both these equations together to get:
2LW + 2 = 132
subtract 2 from both sides of these equations to get:
2LW = 130
divide both sides of these equations by 2 to get:
LW = 65
substitute for LW in the original 2 equations to get:
LW + L + W + 1 = 84
LW - L - W + 1 = 48
become:
65 + L + W + 1 = 84
65 - L - W + 1 = 48
subtract 66 from both sides of these equations to get:
L + W = 18
-L -W = -18
Solve for L in each equation to get:
L = 18 - W (same answer for each equation).
You have a value for L in relation to W.
substitute for L in the first original equation to get:
(L+1)*(W+1) = 84 becomes:
(18-W+1) * (W+1) = 84
simplify each factor to get:
(19-W) * (W + 1) = 84
multiply out the factors to get:
19W + 19 - W^2 - W = 84
combine like terms to get:
-W^2 + 18W + 19 = 84
subtract 84 from both sides of the equation to get:
-W^2 + 18W - 65 = 0
multiply both sides of the equation by -1 to get:
W^2 - 18W + 65 = 0
factor to get:
(W-13) * (W - 5) = 0
you get:
W = 13 or W = 5
since L = 18 - W, then you also get:
When W = 13, L = 5
When W = 5, L = 13
substitute these values in the original equations to get:
When W = 13 and L = 5, then:
(L+1)*(W+1) = 84
(L-1)*(W-1) = 48
become:
6*14 = 84
4*12 = 48
When W = 5 and L = 13, then:
(L+1)*(W+1) = 84
(L-1)*(W-1) = 48
become:
14*6 = 84
12*4 = 48
all equations are true so the values for L and W are good.
Perimeter = 2L + 2W.
When L = 5 and W = 13, perimeter becomes 10 + 26 = 36
When L = 13 and W = 5, perimeter becomes 26 + 10 = 36
your answer is:
perimeter = 36
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