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Hint: This is an area question so proper pictorial representation based upon the data provided in the question will surely help in understanding about the region which we are concerned with. The difference between the area of the square and the area of the circle will take us to the right answer, so use the respective formula for the area to reach the solution.

ABCD is a square whose vertices are given as A (-2, 2), B (2, 2), C (2, -2), D (-2, -2).

In a square of the sides are equal thus AB=BC=CD=DA

Now using the distance formula $D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Using equation (1) we can find out the side AB, so it will be

$AB = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {2 - 2} \right)}^2}} $ As A is (-2, 2) and B is (2, 2)

$ \Rightarrow AB = \sqrt {16 + 0} = \sqrt {16} = 4$ Units

Now as AB=BC=CD=DA=4 units

Now the area of a square is given as, $A = {\left( {side} \right)^2}$â€¦â€¦â€¦â€¦â€¦.. (2)

So side=4 units

Hence putting values in equation (2) we get

$A = {\left( 4 \right)^2} = 16$Unitsâ€¦â€¦â€¦â€¦â€¦â€¦. (3)

Now the circle is centered at origin and its radius is given as r=2cm.

Now the area of circle is given as $a = \pi {r^2}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (4)

Putting the value in equation (3) we get

$a = \pi {(2)^2} = 4\pi c{m^2}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (5)

Clearly,

Required highlighted area = Area of square â€“ Area of circleâ€¦â€¦â€¦â€¦â€¦â€¦â€¦ (5)

So substituting the values from equation (3) and (5) we get

Required highlighted area = $16 - 4\pi {\text{ c}}{{\text{m}}^2}$

Note: Whenever we face such types of problems the key behind the solution lies in the diagrammatic representation of the data provided, then awareness about the basic area formula can help in getting to the right track to reach the answer.

ABCD is a square whose vertices are given as A (-2, 2), B (2, 2), C (2, -2), D (-2, -2).

In a square of the sides are equal thus AB=BC=CD=DA

Now using the distance formula $D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Using equation (1) we can find out the side AB, so it will be

$AB = \sqrt {{{\left( { - 2 - 2} \right)}^2} + {{\left( {2 - 2} \right)}^2}} $ As A is (-2, 2) and B is (2, 2)

$ \Rightarrow AB = \sqrt {16 + 0} = \sqrt {16} = 4$ Units

Now as AB=BC=CD=DA=4 units

Now the area of a square is given as, $A = {\left( {side} \right)^2}$â€¦â€¦â€¦â€¦â€¦.. (2)

So side=4 units

Hence putting values in equation (2) we get

$A = {\left( 4 \right)^2} = 16$Unitsâ€¦â€¦â€¦â€¦â€¦â€¦. (3)

Now the circle is centered at origin and its radius is given as r=2cm.

Now the area of circle is given as $a = \pi {r^2}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (4)

Putting the value in equation (3) we get

$a = \pi {(2)^2} = 4\pi c{m^2}$â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (5)

Clearly,

Required highlighted area = Area of square â€“ Area of circleâ€¦â€¦â€¦â€¦â€¦â€¦â€¦ (5)

So substituting the values from equation (3) and (5) we get

Required highlighted area = $16 - 4\pi {\text{ c}}{{\text{m}}^2}$

Note: Whenever we face such types of problems the key behind the solution lies in the diagrammatic representation of the data provided, then awareness about the basic area formula can help in getting to the right track to reach the answer.

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