LainIwakura
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Pls help ( 45 points )

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 the possible dimension of the floor?

Respuesta :

felixkings

Answer:

[tex]the \: floor \: has \: the \: following \: dimensions \to \\ \boxed{ length \: of \: 37.35 \: meters} \\ and \\ \boxed{width \: of \: 5.35 \: meters}[/tex]

Step-by-step explanation:

[tex]let \: the \: width \: of \: the \: floor \: be \to \: w \\ let \: the \: length \: of \: the \: floor \: be \to \: l = w + 32 \\ but \: the\: are a\: of \: the \: rectangular \: floor \: is \to \\ a = lw = 200 \\ w(w + 32) = 200 \\ {w}^{2} + 32w = 200 \\ {w}^{2} + 32w - 200 = 0 \\ w = \frac{ - 32 + \sqrt{32 {}^{2} - 4(1)( - 200) } }{2(1)} \\ \boxed{ w = 5.35} \\ hence \\ l = 5.35 + 32 \\ \boxed{l= 37.35}[/tex]

♨Rage♨

abidemiokin

The dimension of the floor is 5.35m by 37.35m

The perimeter of the floor is expressed using the formula below:

P = 2(L+W)

L is the length

W is the width

If the length of the floor is 32m longer than its width, then;

L = 32 + W

Since the area is greater than 200 square metres, hence:

A > 200

LW > 200

(32+W)W > 200

32W+W² > 200

W² + 32W - 200 > 0

On factorizing

Width of the floor W = 5.35m

Since the Length = 32 + W

Length = 32 + 5.35

Length = 37.35m

Hence the dimension of the floor is 5.35m by 37.35m

Learn more here: https://brainly.com/question/19088466

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