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]