SOLUTION: Hector paints a picture which is 10 inches longer than it is wide. When he frames it, the outside dimensions (that is, the length and the width) are each 2 inches longer. If the ar
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Question 242382: Hector paints a picture which is 10 inches longer than it is wide. When he frames it, the outside dimensions (that is, the length and the width) are each 2 inches longer. If the area of the picture with the frame is 40 square inches more than the area of the picture without its frame, what is the length of the original painting? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Hector paints a picture which is 10 inches longer than it is wide.
When he frames it, the outside dimensions (that is, the length and the width) are each 2 inches longer.
If the area of the picture with the frame is 40 square inches more than the area of the picture without its frame, what is the length of the original painting?
:
Let x = the width of the picture
then
(x+10) = the length of the picture
:
They say that when framed it adds 2" to each dimension of the picture:
(x+2) = overall width
and
(x+10) + 2 = (x+12) = overall length
:
"the area of the picture with the frame is 40 square inches more than the area of the picture without its frame," therefore:
:
overall area - picture area = 40 sq/in
(x+12)(x+2) - x(x+10) = 40
FOIL
x^2 + 2x + 12x + 24 - (x^2 + 10x) = 40
remove brackets, change signs
x^2 + 14x + 24 - x^2 - 10x = 40
Combine like terms
x^2 - x^2 + 14x - 10x + 24 = 40
:
4x = 40 - 24
:
4x = 16
x =
x = 4 inches is the width of the picture
then
4 + 10 = 14 inches is the length of the picture
:
:
See if that is true, by finding the area of each:
16*6 = 96
14*4 = 56
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diff = 40
