Question

Pls help ( 45 points )

Answer 1

Answer:

$the \: floor \: has \: the \: following \: dimensions \to \\ \boxed{ length \: of \: 37.35 \: meters} \\ and \\ \boxed{width \: of \: 5.35 \: meters}$

Step-by-step explanation:

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

♨Rage♨