SOLUTION: Hello,
I need help with this problem:
The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an
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I need help with this problem:
The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an
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The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an area of 96 square inches. What are the dimensions of the picture? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an area of 96 square inches. What are the dimensions of the picture?
:
Let x = width of the picture
then
(2x-1) = length of the picture
:
It would help to draw a rough diagram of this. Label the picture dimensions as
x and (2x-1). Label the uniform width around the picture as 2 inches. It will be
apparent that the overall dimensions of the frame will be 2x+3, (from 2x-1+4) and x+4.
:
The overall dimensions would be (2x+3) by (x+4), it's area is given as: 96 sq/in
Length time width = area. Therefore:
(2x + 3)*(x+4) = 96
FOIL
2x^2 + 11x + 12 = 96
:
2x^2 + 11x + 12 - 96 = 0
:
2x^2 + 11x - 84 = 0; a quadratic equation
:
Use the quadratic formula to find x: a=2; b=11; c=-84
:
:
: ;minus a minus is a plus
:
: ; we only want the positive solution here
:
:
x = 4.29 inches
:
Find the dimension of the picture:
Length: 2(4.29) - 1 = 7.58 inches
Width = 4.29 inches
:
:
Check our solution by finding the frame area:
(7.58+4) * (4.29+4) =
11.58 * 8.29 = 95.998 ~ 96 sq/in, the given area
:
:
Did this make sense to you? Any questions?
