Question

the length and breadth of rectangular park are 7:5 . a path 1.5 m wide running all around outside the park has an area of 513m² .find perimer of the park​

Answer 1

Step-by-step explanation:

Let the length of the park rectangle ABCD be 8x m

Let the breadth of the park rectangle ABCD be 5x m

Area of rectangle ABCD=(8x×5x)m

=40x

2

m

Length of the park including path PQRS

=8x+2(width of path)

=8x+2(1.5)

=8x+3m

Breadth of the park including path PQRS

=5x+2(width of path)

=5x+2(1.5)

=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=585/39=>x=15

Length of the park=8x

=8×15

=120m

Breadth of the park=5x

=5×15

=75m

Answer 2

Let l=7x

and b= 5x

width of the path = 1.5m

therefore,

Area=35x²

new length= 7x+3m

breadth=5x+3m

Area= (7x+3m)(5x+3m)

35x²+513m²=35x²+21mx+15mx+9m²

513m²=36mx+9m²

513m-9m=36x

504m=36x

x=14m

length=7× 14m=98m

breadth=5×14m= 70m

p=2(l+b)

=2(98m+70m)

=2×168m=336m