A rectangular parcel of land has an area of [tex]7,000 \, \text{ft}^2[/tex]. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot?

let the dimensions of the sides be x and y x y=7000 and sqrt(x^2+y^2)=x+10 (sqrt(x^2+y^2) is how you find the length of the diagonal) =>y=7000/x => sqrt(x^2+(7000/x)^2)=x+10 => x^2+49,000,000/(x^2)=(x+10)^2 => x^2+49,000,000/(x^2)=x^2+20x+100 => x^4+49000000=x^4+20x^3+100x^2 => 20x^3+100x^2-49000000=0 => use your calculator =>x approx = 133 since x y=7000, y=7000/133 => y approx = 53

Answered by paulcox | 2024-06-10

Dimension of land = 133.16 ft x 52.57 ft ;

Answered by agaue | 2024-06-12

The dimensions of the rectangular parcel of land are approximately 133 ft by 53 ft, calculated using the area and the diagonal properties. The width can be found by dividing the area by the length. All calculations took into account the relationships given in the problem statement.
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Answered by paulcox | 2024-10-10