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\(\text{A shed that measures 8 feet by 11 feet has a brick pathway of constant width built around it, so that the outer edge of the pathway is also a rectangle. The total area of the pathway is 120 square feet. What is its width (in feet)? }\)

 Jul 17, 2019
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A shed that measures 8 feet by 11 feet has a brick pathway of constant width built around it, so that the outer edge of the pathway is also a rectangle. The total area of the pathway is 120 square feet. What is its width (in feet)?

The path width is x.

 

\((8+2x)\times (11+2x)-8\times 11=120\\ 88+16x+22x+4x^2-88=120\\ 4x^2+38x-120=0\\ x^2-9,5x-30=0\\ x=4,75\pm\sqrt{4,75^2+30}\\ x=12\)

\(x =-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\)

\(x=4,75\pm\sqrt{4,75^2+120}\\ \color{blue} x=12\)

 

The width is 12ft.

laugh  !

 Jul 17, 2019

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