Question

A rectangular park has a dimension of 14 m by 20 m is surrounded by a pathway of uniform width. If
the area of the pathway is 152 m², find the width of the pathway.

1. What equation represents the problem?
2 What is the width of the pathway?​

Answer 1

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

Answer 2

Given :- A rectangular park has a dimension of 14 m by 20 m is surrounded by a pathway of uniform width. If

the area of the pathway is 152 m², find the width of the pathway.

To Find :-

• 1. What equation represents the problem ?
• 2 What is the width of the pathway ?

Solution :-

we know that,

• Area of rectangle = Length * Breadth .

So ,

→ Area of rectangular park = 14 * 20 = 280 m².

Now, Let us assume that, the width of the pathway is x m.

So,

• Length of Outer rectangle now = Length of park + 2 * width = (14 + 2x) m .
• Breadth of Outer rectangle now = Breadth of park + 2 * width = (20 + 2x) m .

then,

→ Area of Outer rectangle = (14 + 2x)(20 + 2x) = 280 + 28x + 40x + 4x² = (4x² + 68x + 280) m² .

Therefore,

→ Area of pathway = Area of Outer rectangle - Area of rectangular park .

→ (4x² + 68x + 280) - 280 = 152 m² .

→ 4x² + 68x = 152

→ 4x² + 68x - 152 = 0

→ 4(x² + 17x - 38) = 0

→ x² + 17x - 38 = 0 { Answer 1 } .

Now,

→ x² + 17x - 38 = 0

→ x² + 19x - 2x - 38 = 0

→ x(x + 19) - 2(x + 19) = 0

→ (x + 19)(x - 2) = 0

→ x = (-19) or 2 .

since width of path in negative value is not Possible.

Hence, width of the pathway is 2 m. { Answer 2 } .

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