Question

a rectangular field is 16m long and 10m wide. There is a path of uniform width all around it,having an area of 120m square. Find the width of the path.

Answer 1

PQRS= rect field
ABCD= outer rect 

1: length of PQRS= l = PQ= 16m
Breadth = b= QR= 10m
width = xm
length of ABCD = I' = AB= 16 +2x
breadth= b' = CD= 10+ 2x

2: length of rect field PQRS= l= 16
breadth = b = 10
Area of PQRS= l x b = 160
Area of ABCD= l' x b' = (16+2x)(10+2x)= 4x² +52x + 160

3: area of path = area of PQRS = area of ABCD 
= 4x² +52x + 160 - 160 
= 4x²+ 52x

4: Area of path = 120
so, 120 = 4x²+ 52x
x= 2 and x= -15 
-15 will be neglected so x=2

Answer 2

☆ Solution ==》

Answer:-

Width of the path = 2m

Given=》

length = 16m

width = 10m

Area= 120 m²

To find =》

Width =?

Explanation =》

Let the width of the path = x m

Dimensions of rectangle

length = 16m , b= 10m

Dimension of rectangle EFGH

length = ( 10 +2x)

breath = ( 10+2x)

Area of path = 120m²

Area of EFGH - Area of ABCD = Area of path

= ( 16+2x)(10+2x)-16×10=120

= 16(10+2x)+2x(10+2x)-160=120

= 160+32x+200x+4x² -160=120

= 4x²+5x-120=0

= 4( x²+13x-30)=0

= x²+13x-30=0

= x²+(15-2)x-30=0

= x²+15x-2x-30=0

= x(x+15)-2(x+15)=0

= ( x+15)(x-2)

= x= -15( Neglected), x =2

= x=2

Width of the path = 2m