A rectangular parcel of land has an area of 7,000 ft2. 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

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Agaueid Agaueid · Answer 2

Answer:

Dimension of land = 133.16 ft x 52.57 ft

Step-by-step explanation:

Area of rectangle = Length x Breadth = LB = 7000 ft²

Diagonal of rectangle is given by

[tex]d=\sqrt{L^2+B^2}[/tex]

Given that diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel

That is

[tex]\sqrt{L^2+B^2}=B+10[/tex]

Taking square on both sides

[tex]L^2+B^2=B^2+20B+100\\\\L^2=20B+100[/tex]

We also have

LB = 7000

[tex]LB=7000\\\\L=\frac{7000}{B}\\\\L^2=\frac{7000^2}{B^2}\\\\20B+100=\frac{7000^2}{B^2}\\\\20B^3+100B^2=49\times 10^6\\\\B^3+5B^2-2450000=0\\\\B=133.16ft\\\\L=\frac{7000}{133.16}=52.57ft[/tex]

Dimension of land = 133.16 ft x 52.57 ft