Ok so, The Smiths have decided to put a paved walkway of uniform width around their swimming pool. The pool is a rectangular pool that measures 12ft by 20ft. The area of the walkway will be 68 square feet. Find the width of the walkway.
It's part of my review for my test and I don't get what I'm supposed to use except for the quadratic formula.
It's part of my review for my test and I don't get what I'm supposed to use except for the quadratic formula.
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Did you draw a picture?
Draw a rectangle representing the pool. Label the long sides as having length 20ft and the short sides as having length 12ft. Now draw a larger rectangle around this rectangle, a constant distance from all four sides of the smaller rectangle. The region between the two rectangles represents the walkway. Label its constant width as x (ft).
Extend the sides of the smaller rectangle to meet the larger rectangle. Observe that the walkway consists of 4 squares of side x (at the corners), two rectangles that are 20ft by x ft, and two rectangles that are 12ft by x ft. Hence, the area A of the walkway is the sum of these:
A = 4x² + 2(20x) + 2(12x) = 4x² + 64x
This is to equal 68 ft², so
4x² + 64x = 68
4x² + 64x - 68 = 0
To make the numbers easier, divide both sides by 4:
x² + 16x - 17 = 0
This readily factors to (x + 17)(x - 1) = 0, but let's use the quadratic formula:
x = [-16 ± √(16² - (4)(1)(-17))]/((2)(1))
= (-16 ± √324)/2
= (-16 ± 18)/2
= -8 ± 9
So x = 1 or x = -17. x = -17 is unphysical, so we discard that solution. Hence, the walkway is 1 foot wide.
Draw a rectangle representing the pool. Label the long sides as having length 20ft and the short sides as having length 12ft. Now draw a larger rectangle around this rectangle, a constant distance from all four sides of the smaller rectangle. The region between the two rectangles represents the walkway. Label its constant width as x (ft).
Extend the sides of the smaller rectangle to meet the larger rectangle. Observe that the walkway consists of 4 squares of side x (at the corners), two rectangles that are 20ft by x ft, and two rectangles that are 12ft by x ft. Hence, the area A of the walkway is the sum of these:
A = 4x² + 2(20x) + 2(12x) = 4x² + 64x
This is to equal 68 ft², so
4x² + 64x = 68
4x² + 64x - 68 = 0
To make the numbers easier, divide both sides by 4:
x² + 16x - 17 = 0
This readily factors to (x + 17)(x - 1) = 0, but let's use the quadratic formula:
x = [-16 ± √(16² - (4)(1)(-17))]/((2)(1))
= (-16 ± √324)/2
= (-16 ± 18)/2
= -8 ± 9
So x = 1 or x = -17. x = -17 is unphysical, so we discard that solution. Hence, the walkway is 1 foot wide.