Question

help need help need an answer​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that width of rectangular floor be x feet

So, Length of rectangular floor be x + 26 feet

Given that, Area of Rectangular floor should be less than 2040 square feet.

$\rm\implies \:x(x + 26) < 2040$

$\rm\implies \: {x}^{2} + 26x - 2040 < 0$

$\rm \: {x}^{2} + 60x - 34x - 2040 < 0$

$\rm \: x(x + 60) - 34(x + 60) < 0$

$\rm \: (x + 60)(x - 34) < 0$

$\rm\implies \: - 60 < x < 34$

But

$\rm \: x > 0$

$\rm\implies \:0 < x < 34$

On adding 26 in each term, we get

$\rm \: 0 + 26 < x + 26 < 34 + 26$

$\rm \: 26 < x + 26 < 60$

So, it implies

Width lies in the range (0, 34)

and

Length lies in the range (26, 60)

If area of floor is 360 square feet, in that case

$\rm \: x(x + 26) = 360$

$\rm \: {x}^{2} + 26x - 360 = 0$

$\rm \: {x}^{2} + 36x - 10x - 360 = 0$

$\rm \: x(x + 36) - 10(x + 36) = 0$

$\rm \: (x + 36)(x - 10) = 0$

$\bf\implies \:x = 10$

So, in this situation,

Width of rectangular floor = 10 feet

Length of Rectangular floor = 36 feet

It means,it would be realistic to have floor area 360 square feet.

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$