Question

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?

Answer 1

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

Answer 2

Dimension of land = 133.16 ft x 52.57 ft

Explanation:

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

Diagonal of rectangle is given by

`d=√(L^2+B^2)`

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

That is

`√(L^2+B^2)=B+10`

Taking square on both sides

`L^2+B^2=B^2+20B+100\\\\L^2=20B+100`

We also have

LB = 7000

`LB=7000\\\\L=(7000)/(B)\\\\L^2=(7000^2)/(B^2)\\\\20B+100=(7000^2)/(B^2)\\\\20B^3+100B^2=49* 10^6\\\\B^3+5B^2-2450000=0\\\\B=133.16ft\\\\L=(7000)/(133.16)=52.57ft`

Dimension of land = 133.16 ft x 52.57 ft