Question

The length of a rectangle is 5 ft less than twice the width, and the area of the rectangle is 52 ft2. Find the dimensions of the rectangle.

Answer 1

Answer:

$Length=8 feet\\\\Width=6.5 feet$

Step-by-step explanation:

let the width pf the rectangle be 'w' feet

length(L) of the rectangle= $2w-5$

Area of the rectangle= $52 feet ^2$

Area of a rectangle =Length*Width

$52=(2w-5)*w\\\\2w^2-5w=52\\\\2w^2-5w-52=0$

Making the factors of the equation:

$2w^2+8w-13w-52=0$

Taking common from the equation:

$2w(w+4)-13(w+4)=0\\\\(w+4)(2w-13)=0$

$w+4=0\\\\w=-4$

OR

$2w-13=0\\\\2w=13\\\\w=13/2\\\\w=6.5 feet$

As width of the rectangle cannot be negative, so $width(w)= 6.5 feet$

$Length(L)=2w-5=2(6.5)-5 = 13-5=8 feet$

$Length=8 feet\\\\Width=6.5 feet$