Question

The length of the floor is 32 m longer than its width and there is greater than

200 square meters. You will cover the floor completely with tiles. What will be the

possible dimension of the floor​

Answer 1

$\sf{The~floor~has~the~following~dimensions→}\\\longrightarrow\fbox{length~of~37.35~metres}$

and

$\longrightarrow\fbox{width~of~5.35~metres}$

$\sf{The~width~of~the~floor~will~be→w}\\\sf{But~the~area~of~the~rectangular~floor~is→}$

$\sf{a=lw=200}\\\sf{w(w+32)=200}\\\sf{w^2+32w=200}\\\sf{w^2+32w-200=0}\\\sf{w=}\frac{-32+\sqrt{32^2-4(1)(-200)} }{2(1)}$

$\longrightarrow$ $\fbox{5.35+32}$

Hence,

$\sf{l=5.35+32}\\\longrightarrow\fbox{l=37.35}$