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

Given Length of a rectangular hall =5 m more than breadth..

so if Breadth(b) is x ..then length(l) = x + 5

Given area of the hall = 594m^2

we know that Area of a Rectangle..

= l × b

x(x+5) = 594

x^2 + 5x - 594 = 0

x^2 + 27x -22x -594 = 0

x(x+27) -22(x+27)

(x+27)(x-22)

x= -27 or 22.. ignoring negative value...

so x = 22..

so Breadth= 22 and length = 22+5 =27

now Perimeter of a rectangle= 2(L + b)m

=2(22+27)

=2(49)

=98m

Answer 2

Solution:

Consider ABCD is a rectangular hall

Take Breadth = x m

Length = (x + 5) m

We know that

Area of rectangular field = l × b

Substituting the values

594 = x (x + 5)

By further calculation

594 = x2 + 5x

0 = x2 + 5x – 594

x2 + 5x – 594 = 0

It can be written as

x2 + 27x – 22x – 594 = 0

Taking out the common terms

x (x + 27) – 22 (x + 27) = 0

So we get

(x – 22) (x + 27) = 0

Here

x – 22 = 0 or x + 27 = 0

We get

x = 22 m or x = -27 which is not possible

We know that

Breadth = 22 m

Length = (x + 5) = 22 + 5 = 27 m

Perimeter = 2 (l + b)

Substituting the values

= 2 (27 + 22)

By further calculation

= 2 × 49

= 98 m

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