A rectangle garden is 30 ft by 40 ft. Part of the garden is removed in order to install a walkway of uniform width around it. The area of the new garden is one-half the area of the old garden. How wide is the walkway?
Let the walkway have width w
The new dimensions of the garden are
(30-2w) x (40-2w)
That's half of the original area, 30x40
(30-2w)(40-2w) = 1/2 * 30*40
1200 - 140w + 4w^2 = 600
4w^2 - 140w + 600 = 0
w^2 - 35w + 150 = 0
(w-5)(w-30) = 0
w=30 doesn't make much sense. No garden left
w=5 means that the new garden is
20x30 = 1/2 of 30x40
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To find the width of the walkway, we can start by finding the area of the original garden and the area of the new garden.
The area of a rectangle is calculated by multiplying its length by its width.
Given that the original garden is 30 ft by 40 ft, its area can be calculated as:
Original garden area = 30 ft * 40 ft = 1200 square feet
Now, let's represent the width of the walkway by 'x'. Since the walkway is added equally on all sides of the original garden, the dimensions of the new garden would be as follows:
Length: 30 ft + 2x
Width: 40 ft + 2x
The area of the new garden can be calculated by multiplying the length by the width:
New garden area = (30 ft + 2x) * (40 ft + 2x)
The problem states that the area of the new garden is one-half the area of the old garden. Therefore:
New garden area = (1/2) * Original garden area
Substituting the values, we have:
(30 ft + 2x) * (40 ft + 2x) = (1/2) * 1200 square feet
Simplifying the equation:
(30 + 2x) * (40 + 2x) = 600
Now, we can expand and rearrange the equation:
1200 + 60x + 80x + 4x^2 = 600
Combining like terms:
4x^2 + 140x + 1200 - 600 = 0
Simplifying:
4x^2 + 140x - 600 = 0
Dividing the equation by 4 to simplify further:
x^2 + 35x - 150 = 0
Now, we can solve this quadratic equation either by factoring or using the quadratic formula. In this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 1, b = 35, and c = -150. Substituting these values into the quadratic formula:
x = (-35 ± √(35^2 - 4 * 1 * -150)) / (2 * 1)
x = (-35 ± √(1225 + 600)) / 2
x = (-35 ± √(1825)) / 2
x = (-35 ± 42.77) / 2
Now, we have two possible solutions:
x = (-35 + 42.77) / 2 ≈ 3.885
x = (-35 - 42.77) / 2 ≈ -38.885 (since the width cannot be negative)
Therefore, the width of the walkway is approximately 3.885 feet.