Question

The length of the floor is 32m longer than its width and there is greater than 200 square meters. you will cover the floor completely with tiles. what will be 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,

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

Answer 2

The dimensions of the floor are 5.35 x 37.35 $m^{2}$

Step-by-step explanation:

Let us take the width of the floor to be x m. Then, the length of the floor will be= 32 + x

Given area of the floor= 200 sq.m

To find= length and width

Calculations:

Area of a rectangle= l x b

Area of the rectangular floor= (32+x)x

= 32x + $x^{2}$

200 = 32x +  $x^{2}$

$x^{2}$+ 32x - 200

Finding roots of the equation:

x= -b±$\sqrt{b^{2}-4ac }/ 2a$

x= 5.35

So, length= 5.35+ 32= 37.35m

Result:

The dimensions of the floor are 5.35 x 37.35  $m^{2}$

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