Question

2)A park 95 m long and 80 m wide has a path 5 m wide running outside and all around it find the areas of the path
and that of the park.​

Answer 1

Given :

• Length of the park = 95 m
• Breadth of the park = 80 m
• The park is surrounded by a park of width = 5 m

To find :

• Area of the park
• Area of the path

Concept :

To find the area of the park :-

• Use the formula of area of rectangle.

To find the area of the path :-

• Add (5 + 5)m in the dimensions of the park.
• The resultant values will be the dimensions of the park including the path.
• To find the area of the path subtract the area of the park from the area of the park including the path.

Formula to calculate area of rectangle :-

$\boxed{ \pmb{ \sf Area \: \: of \: \: rectangle = l \times b}}$

where,

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

Solution :

Area of the park :-

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: park= l \times b$

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: park= 95 \times 80$

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: park= 7600$

$\\ \dashrightarrow \quad\ \pmb{ \sf Area \: \: of \: \: park = 7600 \: {m}^{2} } \quad \star$

Dimensions of park + path :-

• Length of the park + path = 95 + (5 + 5) = 105 m
• Breadth of the park + path = 80 + (5 + 5) = 90 m

Area of the park + path :-

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: the \: \: park + path = l \times b$

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: the \: \: park + path = 105\times 90$

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: the \: \: park + path = 9450$

$\\ \dashrightarrow \quad\sf \pmb{ Area \: \: of \: \: the \: \: park + path = 9450 \: {m}^{2}}$

→ Area of the path = Area of the park + path - Area of the park

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: the \: \: path = 9450 - 7600$

$\\ \dashrightarrow \quad\sf Area \: \: of \: \: the \: \: path = 1850$

$\\ \dashrightarrow \quad\sf \pmb{Area \: \: of \: \: the \: \: path = 1850 \: {m}^{2}}$

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Know More :

Properties of rectangle :-

• The opposite side of rectangle are parallel to each other.
• The opposite sides of rectangle are equal to each other.
• The diagonals bisect each other.
• The sum of interior angles of rectangle sums upto 360°.

Some related formulae :-

• Perimeter of reactangle = 2(l + b)
• Diagonal of rectangle = √l² + √b²
• Area of square = side × side
• Perimeter of square = 4 × side
• Diagonal of square = √2a
• Total surface area of cube = 6 × side²
• Volume of cube = side³
• Perimeter of cube = 12 × side
• Total surface Area of cuboid = 2(lb + bh + hl)
• Volume of cuboid =  length × breadth × height
• Perimeter of cuboid = 4 (length + breadth + height)

Answer 2

Given :

• Length of park = 95m
• Breadth of park = 80m
• Width of path = 5m
• Path is running outside the park.

To Find :

• Area of park
• Area of path

Solution :

⟹ Area of Inner Park = L × B

⟹ Area of Inner Park = 95 × 80

⟹ Area of Inner Park = 7,600m²

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Now,

✰ Dimensions of Outer park will be given by actual length / breadth + width + width of path . We are adding 10 to the actual length and breadth as it's outside the park.

⟹ Length of Outer Park :

• Length = Actual length + Width
• Length = 95 + 10
• Length = 105m

⟹ Breadth of Outer Park :

• Breadth = Actual breadth + Width
• Breadth = 80 + 10
• Breadth = 90m

⟹ Area of Outer Park :

• Area of Outer park = L × B
• Area of Outer park = 105 × 90
• Area of Outer park = 9,450m²

________________

Now,

✰ Area of Path will be given by Area of Outer Park - Area of Inner Park .

⟹ Area of Path :

• Area = Bigger area - Smaller area
• Area = 9450 - 7600
• Area = 1,850m²

Thus Area of Path is 1,850m²

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