Question

The length of a rectangular hall is 5 m more than its breadth . If the area of the hall is 594 m^2 , find its perimeter.

Answer 1

let breadth=A,

then
length=A+5,

(A+5).A=594,


(A+5).A=27×22,

then
A=22,

hence
breadth=22 me,

length=27 me,

therefore
perimeter=2(27+22),

=2×49,

=98

Answer 2

$\large{\bold{Solution:}}$

.

Let the breadth be $\textbf{x}$.

Thus, length is $\bold{x + 5}$

Area of rectangle = $\bold{Length \times Breadth}$

= > $\textbf{594 = x}( + 5)$

= > $\bold{594 = {x}^{2} + 5x}$

= > $\bold{0 = {x}^{2} + 5x - 594}$

= > $\bold{0 = {x}^{2} + 27x - 22x - 594}$

= > $\bold{0 = x(x + 27) - 22( \times + 27)}$

= > $\bold{0 = (x - 22)(x + 27)}$

= > $\bold{\fbox{x = 22}}$

$\bold{x = 22}$ ( Breadth )

Then,

Length $\bold{22 + 5 = 27}$

Perimeter of rectangle $\bold{= 2(l + b)}$

Perimeter of rectangle $\bold{= 2(22 + 27)}$

$\bold{= 2 \times 49}$

$\bold{\fbox{= 98\:m}}$

$\textbf{\underline{Therefore, perimeter is 98 m.}}$

.

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$\bold{\fbox{Hope\:It\:Helps\:!!!}}$
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