The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions?

1 Answer

build the equations and solve...

Explanation:

let the area be A = l * w where length is l and width is w

so 1.st equqtion will be

l*w=65

and length is 3 yd less than double the width says:

l = 2w-3 (2nd eq.)

replacing l with 2w-3 in first eq. will yield

(2w-3) * w = 65
2w^2-3w = 65
2w^2-3w-65 = 0

now we have a 2nd order equation just find the roots and take the positive one as width can not be negative...

w=(3+-sqrt(9+4*2*65))/(2*2) = (3+-sqrt(529))/(4) = (3+-23)/4
w=-5 , 13/2 so taking w=13/2 = 6.5 yd

replacing w with 6,5 in second eq. we get

l=2w-3 = 2*6.5-3 = 13-3 = 10 yd

A = l*w = 10 * 6.5 = 65yd will confirm us...