SOLUTION: The original problem was:
[A photo is 4 inches longer than it is wide. A 3-inch border is placed around the photo making the total area of the photo and border 165 square inches.
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[A photo is 4 inches longer than it is wide. A 3-inch border is placed around the photo making the total area of the photo and border 165 square inches.
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Question 185501: The original problem was:
[A photo is 4 inches longer than it is wide. A 3-inch border is placed around the photo making the total area of the photo and border 165 square inches. What are the dimensions of the photo? ]
So the initial question was…what was the dimensions of the photo. Since the total area of the photo AND border was 165 square inches, logically the
Total Area of Border and Photo (R3) = 165 square inches or the sum of the Area of the Border (R2) and Photo (R1).
So if I want to find the area of the photo the equation should read R3 - R1 = R2 since the total area of 165 subtracting the area of the photo should allow us to solve for the border and then substitute for w.
Let w = width of photo and w + 4 = length of photo so that the dimensions will equal (w)(w+4) = R1
The area of the border can be seen as the added areas of smaller rectangles surrounding the photo. So that :
x +4 + 6 (3) Each long rectangle = (3x + 30) x 2
3x (x)(x+4) 3x Each smaller rectangle = (3x) x 2
x + 4 + 6 (3)
Therefore the area of the border is represented by 6x + 60 + 6x or 12x + 60
The whole equation can be seen as:
165 – area of photo = area of border or
165 – (x2 + 4x) = 12x + 60, using addition principle combine like terms
165 – x2 - 4x = 12x + 60 becomes 0 = x2 +16x -105
And this is where I get stuck because I cant find the factors of 16 and 105 that make this equation solvable?
Can you please show me the solution to this problem so that I understand.
You can put this solution on YOUR website! Let L = length of photo and W = width of photo.
From the problem... "A photo is 4 inches longer than it is wide"
A 3-inch border is placed around the photo making the total area 165 sq.ins.
The total area of the photo plus 3-inch border can be expressed as: because the addition of the 3-inch border adds a total of 6 inches to the length and 6 inches to the width.
Now Substitute A = 165 and L = W+4. Simplify the left side. Subtract 165 from both sides. Factor this quadratic equation. so that... or Discard the negative solution as lengths are positive quantities.
The width is 5 inches and the length is 9 inches.
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