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
Answers
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
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