A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 $m^{2}$ . Find the width of the path.

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Solution

Area of the outer rectangle

= $(16 + 2 x) (10 + 2 x)$

= $160 + 32 x + 20 x + 4 x^{2}$

= $160 + 52 x + 4 x^{2}$

Area of the inner rectangle

= $16 m \times 10 m$

= $160 m^{2}$

Area of the path

= $160 + 52 x + 4 x^{2} - 160 = 120$

or, $4 x^{2} + 52 x - 120 = 0$

or, $x^{2} + 13 x - 30 = 0$

or, $(x + 15) (x - 2) = 0$

or, $x = - 15, 2$

As width can not be negative x = 2 m

i. e. width of the path = 2 m

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A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m2. Find the width of the path.

A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 $m^{2}$ . Find the width of the path.