PICTURE FRAMING The matting around the picture has a uniform width. How wide is the matting if its area equals the area of the picture? Round to the nearest of an inch.
|---------------------------|
| <-------5in----------> |
| | wide | |
| | | |
| 4in long | |
| | | |
| | | |
| ---------------------- |
|---------------------------|
2w(5+2w) + 4(2w) = 20
10 w + 4 w^2 + 8 w = 20
4 w^2 + 18 w - 20 = 0
2 w^2 + 9 w - 10 = 0
w = [ -9 +/- sqrt(81+80) ]/4
w = [-9 +/- 12.7 ]/4
w = 0.922
check
0.922 * 8 + 0.922*10 + 4*0.922^2 = 19.99 yes
To determine the width of the matting, we need to find the area of the picture and equate it to the area of the matting.
The area of the picture can be calculated by multiplying its length by its width:
Area of picture = length × width = 4 in × 5 in = 20 in².
Let's denote the width of the matting as x inches.
The outer dimensions of the framed picture, including the matting, can be calculated by increasing the length and width of the picture by twice the width of the matting:
Length of framed picture = 4 in + 2x
Width of framed picture = 5 in + 2x
The area of the framed picture, including the matting, can be calculated by multiplying the outer dimensions:
Area of framed picture = (4 in + 2x) × (5 in + 2x)
Since we are given that the area of the matting equals the area of the picture, we can set up the equation:
Area of framed picture = Area of picture
(4 in + 2x) × (5 in + 2x) = 20 in²
Simplifying the equation, we get:
(4 + 2x)(5 + 2x) = 20
Expanding the equation, we get a quadratic equation:
20 + 8x + 10x + 4x² = 20
4x² + 18x + 20 = 20
4x² + 18x = 0
Dividing the equation by 2, we get:
2x² + 9x = 0
Factoring out an x from the equation, we get:
x(2x + 9) = 0
Setting each factor equal to zero, we find two possible solutions:
x = 0 or 2x + 9 = 0
The width of the matting can't be zero, so we disregard the solution x = 0.
To find the actual value of x, we solve the equation 2x + 9 = 0:
2x = -9
x = -9/2
However, since the width can't be negative (as it represents a physical dimension), we disregard this solution as well.
Therefore, there is no valid solution for the width of the matting that would make its area equal to the area of the picture.