Q3 of 38 Page 5

There is a path of 3m width all around outside the square shaped park of Ajanta Housing Complex. The perimeter of the park including the path is 484 m. Let’s calculate the area of the path.

We have


The Ajanta Housing Complex is ABCD, which is square shaped.


The path around it is 3 m wide uniformly.


We can conclude that,


If ABCD is a square, then


EFGH is also a square.


In ABCD:


Let AB = p


Then, AB = BC = CD = DA = p [∵ all sides are equal in a square]


In EFGH:


EF = AB + 3 + 3 [∵, 3 m width is around the square park]


⇒ EF = AB + 6


⇒ EF = p + 6


Then, EF = FG = GH = HE = p + 6 [∵all sides are equal in a square]


Given that,


Perimeter of park including the path = 484 m


Or


Perimeter of ABCD + Perimeter of EFGH = 484 m


⇒ (4 × AB) + (4 × EF) = 484 [∵, Perimeter of a square = 4 × length]


⇒ (4 × p) + (4 × (p + 6)) = 484


⇒ 4p + 4p + 24 = 484


⇒ 8p + 24 = 484


⇒ 8p = 484 – 24


⇒ 8p = 460




⇒ p = 57.5 m


Thus, AB = BC = CD = DA = 57.5 m


And, EF = FG = GH = HE = 57.5 + 6 = 63.5 m


We need to find the area of the path.


Area of the path = Area of EFGH – Area of ABCD …(A)


Area of EFGH:


Area of square is given as,


Area = (length)2


⇒ Area of EFGH = (EF)2


⇒ Area of EFGH = (63.5)2 …(i)


Area of ABCD:


⇒ Area of ABCD = (AB)2


⇒ Area of ABCD = (57.5)2 …(ii)


Substitute equations (i) and (ii) in (A), we get


Area of the path = (63.5)2 – (57.5)2


⇒ Area of the path = 4032.25 – 3306.25


= 726 m2


Hence, the area of the path is 726 m2.


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