Math, asked by daasjhutha, 17 hours ago

the length and the breadth of a rectangular park are in the ratio 7:5 , a path 1.5m wide , running around outside the park has an area of 513m, find the perimeter of the park​

Answers

Answered by Mïƭsυɾï
8

Define x:

Given that the ratio of length and breadth = 8 : 5

Let the length of the park be 8x

And the breadth of the park be 5x

Find the area of the park in term of x:

Area = Length x Breadth

Area = (8x) x (5x)  = 40x² m²

Find the Length and Breadth of the park and the path:

Length = 8x + 1.5 + 1.5 =  ( 8x + 3 ) m²

Breadth = 5x + 1.5 + 1.5 = (5x + 3) m²

Find the area of the park and the path:

Area = Length x Breadth

Area = (8x + 3) (5x + 3)  m²

Solve x:

Given that the area of the path is 594 m²

(8x + 3) (5x + 3)  - 40x² = 594

40x² + 24x + 15x + 9 - 40x² = 594

39x + 9 = 594

39x = 585

x = 15 m

Find the dimension of the park:

Length = 8x = 8(15) = 120 m

Breadth = 5x = 5(15) = 75 m

Answer: The park is 120 m by 75 m

Answered by llParkMinniell
2

Answer:

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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.

Medium

Solution

verified

Verified by Toppr

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=

39

585

=>x=15

Length of the park=8x

=8×15

=120m

Breadth of the park=5x

=5×15

=75m

hence, answer is 75m

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