Question

The length and breadth of a rectangular park are in the ratio 8:5. A path 1.5 m wide running all round the outside of the park has an area of 594 m${}^2$. Find the dimensions of the park.

Answer 1

Let the length of the park rectangle ABCD be $8xm$

Let the breadth of the park rectangle ABCD be $5xm$

Area of rectangle ABCD=$(8x\times 5x)$m
=$40x^{2}m$

Length of the park including path PQRS
=$8x+2$(width of path)
=$8x+2\left(1.5\right)$
=$8x+3m$

Breadth of the park including path PQRS
=$5x+2$(width of path)
=$5x+2\left(1.5\right)$
=$5x+3m$

Area of the park including path PQRS
=$(8x+3)(5x+3)$
=$(40x^2+39x+9)m^2$
Given,
Area of the path=$594m^2$

Area of PQRS-Area of ABCD=$594$
$=>40x^2+39x+9-40x^2=594$
$=>39x=594-9$
$=>39x=585$
$=>x=\frac{585}{39}$
$=>x=15$

Length of the park=$8x$
=$8\times 15$
=$120m$
Breadth of the park=$5x$
=$5\times 15$
=$75m$