Question

A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path

Answer 1

Given :-

• Length of a rectangular park = 45 m
• Breadth of a rectangular park = 30 m
• Width of a path constructed outside the park = 2.5 m

To find :-

• Area of the path

Knowledge Required :-

→ Formula of area of rectangle :-

• Area of rectangle = l × b

where,

• l = length of the rectangle
• b = breadth of the rectangle

~Understanding the question –

Firstly, we will find the area of the rectangular park then add (2.5 + 2.5) m in the dimensions of the rectangular park that will be the dimensions of the rectangular park including the path. Then find the area of the rectangular park including the path. To find the area of the path subtract the area of the rectangular park from the area of the rectangular park including the path.

Solution :-

Area of the rectangular park :-

⇒ Area = l × b

⇒ Area = 45 × 30

⇒ Area = 1350

Area of the rectangular park = 1350 m²

Dimensions of the rectangular park including the path :-

→ Length = 45 + 2.5 + 2.5 = 50 m

→ Breadth = 30 + 2.5 + 2.5 = 35 m

Area of the rectangular park including the path :-

⇒ Area = l × b

⇒ Area = 50 × 35

⇒ Area = 1750

Area of the rectangular park including the path = 1750 m²

Area of the path = Area of the rectangular park including the path – Area of the rectangular park

→ Area of the path = 1750 - 1350

→ Area of the path = 400

★ Area of the path = 400 m²

Answer 2

$\pmb{\sf {\underline{★ \; Given :}}}$

• Length of park = 45m
• Breadth of park = 30m
• Width of park = 2.5m

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$\pmb {\sf {\underline {★ \; To \; find :}}}$

• Area of path

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$\pmb {\sf {\underline { ★ \; Solution :}}}$

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$\pmb { \sf {Area \; of \; park \; ABCD \; = Length \times Breadth}}$

$\implies \sf { Area \; = 45 \times 30}$

$\implies \sf { Area \; = 1350 \; m^2 }$

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➱ Now, let's find the area of rectangle EFGH ,which is equal to area of park + area of path.

So,

• New length = 45 + 2.5+2.5 = 50m
• New breadth = 30 + 2.5 + 2.5 = 35m

$\pmb {\sf {Area \; of \; rectangle \; EFGH \; = length \times breadth }}$

$\implies \sf { Area \; = 50 \times 35 }$

$\implies \sf { Area \; 1750 \; m^2 }$

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➠ Area of path = Area of EFGH - Area of ABCD

➭ Area of path = 1750 - 1350

➭ Area of path = 400 m²