Question

⭐A rectangular field is 20m long and 14m wide. there is a path of equal width all around it.

Having an area of 111 Sq. m. Find the width of the path.

(HINT:Area of path=area of the field including the path - Area of the field excluding the path) ⭐

CLASS 10 CHAPTER :QUADRATIC EQUATIONS

BEST ANSWER WILL BE MARKED AS BRAINLIEST!

Answer 1

$<b>Hello Friend$

Here is your answer

Given :

↪ Length of a rectangular field = 20 m

↪ Breadth of a rectangular field = 14 m

↪ Area of rectangular field = length × breadth

↪ Area of rectangular field = 20 m × 14 m

↪ Area of rectangle field = 280 m^2

↪ Let the width of the path be x m

Then,

↪ Length of the rectangle formed with the path = ( 20 + x + x ) m

↪ Length of the rectangle formed with the path = ( 20 + 2x ) m

↪ Breadth of the rectangle formed with the path = ( 14 + x + x ) m

↪ Breadth of the rectangle formed with the path = ( 14 + 2x ) m

↪ Area of the rectangle formed with the path = length × breadth

↪ Area of the rectangle formed with the path = ( 20 + 2x ) (14 + 2x )

↪ Area of the rectangle formed with the path = 280 + 40x + 28x + 4x^2

↪ Area of the rectangle formed with the path = 280 + 68x + 4x^2

Now,

↪ Area of rectangle with path - Area of rectangular field = Area of path

$\mathsf {\implies 280 + 68x + 4x^2 - 280 = 111}$

$\mathsf {\implies 68x + 4x^2 - 111 = 0}$

$\mathsf {\implies 4x^2 + 74x - 6x - 111 = 0}$

$\mathsf {\implies 2x ( 2x + 37 ) - 3 ( 2x + 37 ) = 0}$

$\mathsf {\implies ( 2x - 3 ) ( 2x + 37 ) = 0}$

$\mathsf {\implies x = {\dfrac {3}{2}} or x = {\dfrac {-37}{2}}}$

↪ Since width cannot be negative

$\mathsf {Width \: of \: the \: path = 1.5 \: m}$

HOPE IT HELPS

#BE BRAINLY

Answer 2

Answer:

I hope this will help you

Thank you ❤️