A picture is four inches wide and six inches long. It has an additional border of uniform width. If the area of the picture is equal to the area of the border, what is the width of the border?
To find the width of the border, we first need to determine the area of the picture and the area of the border.
Let's start by finding the area of the picture. The picture is given as four inches wide and six inches long. Recall that the area of a rectangle is calculated by multiplying the width by the length.
Area of the picture = Width × Length
= 4 inches × 6 inches
= 24 square inches
Now, we know that the area of the border is equal to the area of the picture. Let's assume the width of the border as "x" inches.
Area of the border = Width of the border × Length of the border
= (4 + 2x) inches × (6 + 2x) inches
= 24 square inches
To solve this equation, we can equate the area of the border to the area of the picture:
24 square inches = (4 + 2x) inches × (6 + 2x) inches
Simplifying the equation, we have:
24 = 24 + 16x + 12x + 4x^2
Rearranging and combining like terms:
4x^2 + 28x = 0
Divide both sides by 4 to simplify:
x^2 + 7x = 0
Now, we can solve the equation by factoring:
x(x + 7) = 0
If we set each factor equal to zero, we have two possible solutions:
x = 0 or x + 7 = 0
The solution x = 0 doesn't make sense in this context because it would mean there is no border. So, we discard that solution.
The remaining solution is x = -7, but since we are dealing with measurements, a negative width doesn't make sense either.
Therefore, there is no valid width for the border in this scenario.
(4+2w)(6+2w) = 2*24 = 48
24 + 20 w + 4 w^2 = 48
4 w^2 + 20 w - 24 = 0
w^2 + 5 w - 6 = 0
(w+6)(w-1) = 0
w = 1
check
4+2 = 6
6+2 = 8
6*8 = 48
48 - 24 = 24 yes