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
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
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
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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