Question

4. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all
around it having an area of 111 sq. metres. Find the width of the path on the outside.

Answer 1

ABCD rectangular field dimensions are,

length ( l ) = 20 m
breadth ( b ) = 14 m

Let the width of the path( w ) = x m

Dimensions of the EFGH rectangle is

length ( L ) = l + 2w

L = ( 20 + 2x )m

Breadth = b + 2w

B = ( 14 + 2x ) m

according to the problem given ,

area of the path = 111 m²

EFGH area - ABCD area = 111 m²

( 20 + 2x )( 14 + 2x ) - 20 × 14 = 111

280 + 40x + 28x + 4x² - 280 - 111 = 0

4x² + 68x² - 111 = 0

splitting the middle term ,

4x² + 74x - 6x - 111 = 0

2x( 2x + 37 ) - 3( 2x + 37 ) = 0

( 2x + 37 )( 2x - 3 ) = 0

2x + 37 = 0 or 2x - 3 = 0

2x = -37 or 2x = 3

x = -37/2 or x = 3/2

width of the path should not be negative .

Therefore ,

width of the path ( w ) = 3/2 m = 1.5 m

I hope this helps you.

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