Arectangle field has an area of 1764 m^2. the width of the field is 13m more than the length. what is the perimeter of the field?

Step-by-step explanation:

hello :

let y the length and  x the  width

A  the area and  P the perimeter

so : A= xy and P  = 2(x+y) y > x and x >0  ; y>0

xy = 1764 ...(1)

y = x +13(2)  put this value in (1) : x (x+13) = 1764

x²+13x -1764 = 0   ...quadratic equation

calculate : Δ = b² -4ac   a =1  and  b =13  and

c = -1764

Δ = (13)² - 4(1)(-1764) = 169 +7056 = 7225=85²

X1 =(-b-√Δ)/2a =(-13-85)/2 < 0 refused  (x  >0)

X2 =(-b+√Δ)/2a =(-13+85)/2  = 72/2 = 36

so : x= 36  put this value in (2) : y = 36+13 = 49

the perimeter is : P  = 2(x+y)

                            P  = 2(36+49) = 170 m