Question

2. A 3 m wide path runs outside and around a rectangular park of length 125 m and
breadth 65 m. Find the area of the path.
and the area

Answer 1

Given

• 3 m wide path runs outside a rectangular path.
• Length of rectangular park is 125 m
• Breadth of rectangular park is 65 m

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To Find

• The area of the path

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Solution

Area of the path = Area of the rectangular park with the path - Area of the rectangular park without the path

So, let's first find the area of the rectangular park without the path.

Area of rectangle ⇒ Lenght × Breadth

Area of rectangle ⇒ 125 × 65

Area of rectangle ⇒ 8125 m²

∴ The area of the rectangular park without the path is 8125 m²

Now let's find the area of the rectangular park with the path.

So we'll add 3 to all four sides of the rectangle. So we'll add 6 to both the length and breadth.

Length ⇒ 125 + 6 = 131 m

Breadth ⇒ 65 + 6 = 71 m

Area of Rectangle ⇒ Length × Breadth

Area of Rectangle ⇒ 131 × 71

Area of Rectangle ⇒ 9301 m²

∴ The area of the rectangular park with the path is 9301 m²

Now let's find the area of the path.

Area of the path = Area of the rectangular park with the path - Area of the rectangular park without the path

Area of path ⇒ 9301 - 8125

Area of path ⇒ 1176 m²

∴ The area of the path is 1176 m²

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Answer 2

Given :-

• Breadth of the path = 3 m
• Length of the rectangular park = 125 m
• breadth of the rectangular park = 65 m

To Find :-

• area of the path runs outside the park ?

Solution :-

• Refer the attached diagram for measuring

• Inner rectangle = park

• outer rectangle = park with path

we know that,

$\boxed{ \orange{ \bold{area \: of \: rectangle = l \times b}}}$

$\bold{: \implies 65m \times 125m} \\ \\ \bold{: \implies 8125 \: {m}^{2} } \: \: \: \: \: \: \: \: \:$

so, area of the park or inner rectangle is 8125 m²

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length of outer rectangle = SR

➪ SR = 3 + DC + 3

➪ SR = 3 + 125 + 3 = 131 m

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ __________________________

Breadth of outer rectangle = SP

➪ SP = 3 + AD + 3

➪ SP = 3 + 65 + 3 = 71m

• Length of outer rectangle = 131m
• breadth of outer rectangle = 71m

then, area of the outer rectangle PQRS

$\bold{: \implies131 m\times 71m} \\ \\ \bold{: \implies9301 \: {m}^{2} } \: \: \: \: \: \: \: \: \:$

so, the area of the outer rectangle is 9301 m²

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now we have,

• ar(PQRS) = 9301 m²
• ar(ABCD) = 8125 m²

$\boxed{\bold{area \: of \: path = ar(PQRS) - ar(ABCD)}}$

$\bold{: \implies 9301 {m}^{2} - 8125 {m}^{2} } \\ \\ \bold{: \implies 1176 \: {m}^{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

hence, the area of the path is 1176 m²