Question

A path all around inside of a
rectangular park 37 metres by 30
metres occupies 570 sq. m, then
the width of the path is​

Answer 1

Given,

Length = 37 m

Breadth = 30m

Area of outer rectangle = 37×30

=1110m

Area of inner rectangle=(area of outer rectangle) - ( area of path)

= 1110 - 570

= 540m²

Let width of path be x then,

Length of inner rectangle=(37-2x)

Breadth of inner rectangle=(30-2x)

Area of inner rectangle= L×b

=(37-2x)(30-2x)

=37(30-2x)-2x(30-2x)

=1110-74x-60x+4x²

=4x²-134x+1110

Now,

4x²-134x+1110=540

Or,4x²-134x+1110-540=0

Or,4x²-134x+570=0

Now,

(Middle term split)

4x²-134x+570

=4x²-20x-114x+570

=4x(x-5)-114(x-5)

=(x-5)(4x-114)

x= 5 or, x = 114/4

So the width is 5 m

Ans

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Answer 2

Given :   A path all around inside of a rectangular park 37 metres by 30

metres occupies 570 sq. m

To find : Width of the Path

Solution:

Rectangular Park  37 m by 30 m

Area of Park = 37 * 30  =  1100 m²

Area of path  = 570  m²

Area of Park excluding Path = 1110 - 570  = 540 m²

Let say Path Width = x

Then  Dimensions of area Excluding path would be  

(37 - 2x)  m  & (30 -2x) m

Area = (37 - 2x)(30 - 2x)   =  1110 + 4x² -134x

1110 + 4x² -134x = 540

=>  4x² -134x  + 570 = 0

=>  2x² - 67x  + 285 = 0

=> 2x² - 10x - 57x + 285 = 0

=> 2x(x - 5) - 57(x - 5) = 0

=> (2x - 57)(x - 5) = 0

=> x = 28.5  or x  = 5

x = 28.5 not possible as then 37 - 2x & 30 - 2x would be - ve

Width of path = 5 m

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