The length of a rectangle is 5ft more than twice its width, and the area of the rectangle is 88ft. How do you find the dimensions of the rectangle?

Let the length & width be `l` feet, & `w` feet, rep.

By what is given, `l=2w+5................(1).`

Next, using the formula : Area of rectangle = length `xx` width, we get another eqn.,
`l*w=88,` or, by `(1)`, `(2w+5)*w=88,` i.e., `2w^2+5w-88=0.`

To factorise this, we observe that `2*88=2*8*11=16*11`, & `16-11=5`.

So we replace, `5w` by `16w-11w`, to get,

`2w^2+16w-11w-88=0.`

`:. 2w(w+8)-11(w+8)=0.`

`:. (w+8)(2w-11)=0.`

`:. w` = width`=-8,` which is not permissible, `w=11/2.`

Then `(1)` gives, `l=16.`

It is easy to verify that the pair `(l,w)` satisfies the given conds.

Hence the dimensions of the rectangle are length`=16` feet, width`=11/2` feet.

The area of the rectangle should be `88 sq.ft` in place of `88` ft mentioned in the question.
Let the width of the rectangle be `x :.` the length will be `2x+5 :.`Area of rectangle is `(2x+5)*x =88 or 2x^2+5x-88=0 or 2x^2+16x-11x-88=0 or 2x(x+8)-11(x+8) =0 or (2x-11)(x+8)=0 :. x= 5.5 or x=-8` Width cannot be negative So `x=5.5 ; 2x+5=16` Hence length is `16ft` and Width is `5.5`ft[Ans]