A rectangular field has an area of 8000 m² and is enclosed by 400 m of fencing. Determine the dimensions to the nearest tenth of a meter.

field:x\ \times\ y\\\\ \left\{\begin{array}{ccc}x\cdot y=8000\\2x+2y=400&/:2\end{array}\right\\\left\{\begin{array}{ccc}x\cdot y=8000\\x+y=200&\to x=200-y\end{array}\right\\\\substitute\ to\ x\cdot y=8000\\\\(200-y)\cdot y=8000\\\\-y^2+200y-8000=0\\\\\Delta=b^2-4ac;\ \Delta=200^2-4\cdot(-1)\cdot(-8000)=40000-32000=8000\\\\y_1=\frac{-b-\sqrt\Delta}{2a};\ y_2=\frac{-b+\sqrt\Delta}{2a}
Δ ​ = 8000 ​ = 1600 ⋅ 5 ​ = 40 5 ​ y 1 ​ = 2 ⋅ ( − 1 ) − 200 − 40 5 ​ ​ = − 2 − 200 − 40 5 ​ ​ = 100 + 20 5 ​ ≈ 144.7 ( m ) y 2 ​ = 2 ⋅ ( − 1 ) − 200 + 40 5 ​ ​ = − 2 − 200 + 40 5 ​ ​ = 100 − 20 5 ​ ≈ 55.3 ( m ) x 1 ​ ≈ 200 − 144.7 = 55.3 ( m ) x 2 ​ ≈ 200 − 55.3 = 144.7 ( m ) A n s w er : D im e n s i o n o f t h e f i e l d i s : 55.3 m × 144.7 m

Answered by Anonymous | 2024-06-10

The dimensions of the rectangular field are calculated using the area and the length of the fencing. We can form equations using these elements and solve to find the dimensions to the nearest tenth of a metre. ;

Answered by JohnRogerStephens | 2024-06-18

The dimensions of the rectangular field are approximately 55.3 meters by 144.7 meters. This is based on its area of 8000 m² and a perimeter of 400 m. By solving a system of equations derived from the area and perimeter, we determined the lengths.
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Answered by Anonymous | 2024-10-12