Question

a rectangular field is 20m long and 14m wife. There is a path of equal width all around it, having an area of 111 m^2. Find the width of the path

Answer 1

Check the pic, i've solved it.

Answer 2

Answer:

1.5 m

Step-by-step explanation:

Length of rectangular field = 20 m

Breadth of rectangular field = 14 m

Let the width of path around field be x

So, length of field including width = 20 +2x

Breadth of field including width = 14 +2x

Now we are  given  that There is a path of equal width all around it, having an area of 111 sq.m.

Area of path = Outer area - Inner Area

$111=(20 +2x)(14 +2x)-(20 \times 14)$

$111=280+40x+28x+4x^2-280$

$111=68x+4x^2$

$68x+4x^2-111=0$

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

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

$x = \frac{-37}{2},\frac{3}{2}$

$x =-18.5,1.5$

Since width cannot be negative

So, Width of path is 1.5 m

Hence the width of the path is 1.5 m

Answer 3

Answer:

1.5 m

Step-by-step explanation:

Length of rectangular field = 20 m

Breadth of rectangular field = 14 m

Let the width of path around field be x

So, length of field including width = 20 +2x

Breadth of field including width = 14 +2x

Now we are  given  that There is a path of equal width all around it, having an area of 111 sq.m.

Area of path = Outer area - Inner Area

$111=(20 +2x)(14 +2x)-(20 \times 14)$

$111=280+40x+28x+4x^2-280$

$111=68x+4x^2$

$68x+4x^2-111=0$

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

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

$x = \frac{-37}{2},\frac{3}{2}$

$x =-18.5,1.5$

Since width cannot be negative

So, Width of path is 1.5 m

Hence the width of the path is 1.5 m