SOLUTION: A framed picture is 7 inches longer than wide. The width of frame is 5 inches. Total area is 470 square inches. Determine outside dimensions of framed picture.
Area = length
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Question 1092130: A framed picture is 7 inches longer than wide. The width of frame is 5 inches. Total area is 470 square inches. Determine outside dimensions of framed picture.
Area = length * width
w = width
l = w + 7
Not sure how to include frame width.
(w + 7)(w) = 470
w^2 + 7w = 470
Not sure from here. Non-homework.
Found 4 solutions by josgarithmetic, ankor@dixie-net.com, ikleyn, greenestamps:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
w, JUST the picture width
L, just the picture length
5, uniform width of frame surrounding picture
Total area frame plus picture, 470 square inches
(for the picture alone; not yet of the frame...)
.
.
.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
L = the length of the picture
w = the width of the picture
:
A framed picture is 7 inches longer than wide.
L = w + 7
The width of frame is 5 inches. Total area is 470 square inches
This would add 10 inches to both length and width
(L+10)*(w+10) = 470
Replace L with (w+7)
((w+7)+10)*(w+10)
(w+17)*(w+10) = 470
FOIL
w^2 + 10w + 17w + 170 = 470
w^2 + 27w + 170 - 470 = 0
w^2 + 27w - 300 = 0
Using the quadratic formula, the positive solution
w = 8.46 in is the width of the picture
then
L = 8.46 + 7 = 15.46 in is the length of the picture
" Determine outside dimensions of framed picture.
Add 10" to each dimension
25.46" by 18.46" the overall dimensions
:
:
:
Check 25.46 * 18.46 = 469.99 ~ 470
Answer by ikleyn(52790) (Show Source): You can put this solution on YOUR website!
.
The formulation of this problem DOES NOT CORRESPOND to its meaning,
which is BAD.
Moreover, my impression is that this discrepancy was made intently,
wich is even worst.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
Since the problem says the total area is 470 square inches, instead of APPROXIMATELY 470 square inches, I would assume the dimensions of the picture are whole numbers.
With a 5 inch border, the dimensions of the picture plus frame are (length of picture) plus 10 and (width of picture) plus 10.
Since we want whole number measurements, we look for ways to write 470 as the product of two whole numbers, each of which must be at least 10.
But the only way to do that is 470 = 47*10, which means the dimensions of the picture itself would be 37 by 0 -- which makes no sense.
So I suspect there is a problem with the numbers you show for the problem.
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