Question

The length and the breadth of a rectangular field are 40 m and 30 m, respectively. It is surrounded on its outside by a uniformly broad path of 2.5 m. What is the are of the path in square metre?

Answer 1

✬ Area of Path = 375 m² ✬

Step-by-step explanation:

Given:

• Length and Breadth of the rectangular field is 40m and 30 m respectively.
• Width of Park which is outside the field is 2.5 m

To Find:

• What is the area of path ?

Solution: Let ABCD is a rectangular park inside the path.

★ Area of rectangle ABCD = Length x Breadth

$\small\implies{\sf }$ Area of ABCD = 40 x 30

$\small\implies{\sf }$ 1200 m²

★ In rectangle EFGH ★

• Length = EF = AB + ( 2.5 + 2.5) = 40+5 = 45 m
• Breadth = FG = BC + (2.5 + 2.5) = 30+5 = 35 m

∴ Area of rectangle EFGH = Length x Breadth = EF x FG

$\small\implies{\sf }$ Area of EFGH = 45 x 35

$\small\implies{\sf }$ 1575 m²

Now we have to find the area of path which lies between the area of EFGH and ABCD. So after subtracting the area of ABCD from EFGH we will get the area of the path .

$\small\implies{\sf }$ Area of path = Ar.( EFGH – ABCD ) m²

$\small\implies{\sf }$ ( 1575 – 1200 ) m²

$\small\implies{\sf }$ 375 m²

Hence, The area of path is 375 m².

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given:

Length and Breadth of the rectangular field is 40m and 30 m respectively.

Width of Park which is outside the field is 2.5 m

To Find:

What is the area of path ?

Solution: Let ABCD is a rectangular park inside the path.

★ Area of rectangle ABCD = Length x Breadth

\small\implies{\sf }⟹ Area of ABCD = 40 x 30

\small\implies{\sf }⟹ 1200 m²

★ In rectangle EFGH ★

Length = EF = AB + ( 2.5 + 2.5) = 40+5 = 45 m

Breadth = FG = BC + (2.5 + 2.5) = 30+5 = 35 m

∴ Area of rectangle EFGH = Length x Breadth = EF x FG

\small\implies{\sf }⟹ Area of EFGH = 45 x 35

\small\implies{\sf }⟹ 1575 m²

Now we have to find the area of path which lies between the area of EFGH and ABCD. So after subtracting the area of ABCD from EFGH we will get the area of the path .

\small\implies{\sf }⟹ Area of path = Ar.( EFGH – ABCD ) m²

\small\implies{\sf }⟹ ( 1575 – 1200 ) m²

\small\implies{\sf }⟹ 375 m²

Hence, The area of path is 375 m².