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Hint: Here, we will find the area of the path by subtracting the inner area from the outer area of the garden.

Here the garden is in the shape of a rectangle.

Given,

Inner length of the garden (AB) $l = 90$ m

Inner width of the garden (BC) $b = 75$ m

As we know that ${\text{Area of a rectangle}} = \left( {{\text{Length of the rectangle}}} \right) \times \left( {{\text{Width of the rectangle}}} \right)$

So, ${\text{Area of inner rectangle ABCD}} = l \times b = 90 \times 75 = 6750$ m2

According to the problem, a path of 5 m wide is to be built around this garden as shown in figure

Here, Outer length of the garden (PQ) $L = l + 5 + 5 = 90 + 5 + 5 = 100$ m

Outer width of the garden (QR) \[B = b + 5 + 5 = 75 + 5 + 5 = 85\] m

So, ${\text{Area of the outer rectangle PQRS}} = L \times B = 100 \times 85 = 8500$ m2

$

{\text{Area of the path}} = {\text{Area of rectangle PQRS}} - {\text{Area of rectangle ABCD}} \\

\Rightarrow {\text{Area of the path}} = 8500 - 6750 = 1750\,{{\text{m}}^2} \\

$

Therefore, the area of the path is 1750 m2.

Since, we know that $1{\text{ hectare}} = 10000{\text{ }}{{\text{m}}^2}$

Also, ${\text{Area of the garden}} = {\text{Area of inner rectangular ABCD}} = 6750{\text{ }}{{\text{m}}^2} = \dfrac{{6750}}{{10000}} = 0.675{\text{ hectare}}$.

Therefore, the area of the garden in hectare is 0.675 hectare.

Note: In the above problem, when a path of 5 m wide is built, the outer dimensions are increased by 10 m in both x and y direction which is clearly visible from the figure.

Here the garden is in the shape of a rectangle.

Given,

Inner length of the garden (AB) $l = 90$ m

Inner width of the garden (BC) $b = 75$ m

As we know that ${\text{Area of a rectangle}} = \left( {{\text{Length of the rectangle}}} \right) \times \left( {{\text{Width of the rectangle}}} \right)$

So, ${\text{Area of inner rectangle ABCD}} = l \times b = 90 \times 75 = 6750$ m2

According to the problem, a path of 5 m wide is to be built around this garden as shown in figure

Here, Outer length of the garden (PQ) $L = l + 5 + 5 = 90 + 5 + 5 = 100$ m

Outer width of the garden (QR) \[B = b + 5 + 5 = 75 + 5 + 5 = 85\] m

So, ${\text{Area of the outer rectangle PQRS}} = L \times B = 100 \times 85 = 8500$ m2

$

{\text{Area of the path}} = {\text{Area of rectangle PQRS}} - {\text{Area of rectangle ABCD}} \\

\Rightarrow {\text{Area of the path}} = 8500 - 6750 = 1750\,{{\text{m}}^2} \\

$

Therefore, the area of the path is 1750 m2.

Since, we know that $1{\text{ hectare}} = 10000{\text{ }}{{\text{m}}^2}$

Also, ${\text{Area of the garden}} = {\text{Area of inner rectangular ABCD}} = 6750{\text{ }}{{\text{m}}^2} = \dfrac{{6750}}{{10000}} = 0.675{\text{ hectare}}$.

Therefore, the area of the garden in hectare is 0.675 hectare.

Note: In the above problem, when a path of 5 m wide is built, the outer dimensions are increased by 10 m in both x and y direction which is clearly visible from the figure.

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