Question
Answers

A square ABCD and a circle are drawn on a graph sheet such that the vertices of the square ABCD are A (-2, 2), B (2, 2), C (2, -2), D (-2, -2) and the circle with center (0, 0) and radius 2 cm is cut off from the above square ABCD. Find the area of the remaining region in the square ABCD.

Answer
VerifiedVerified
148.8k+ views
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}} $…………………… (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, $A = {\left( {side} \right)^2}$…………….. (2)
So side=4 units
Hence putting values in equation (2) we get
$A = {\left( 4 \right)^2} = 16$Units………………. (3)
Now the circle is centered at origin and its radius is given as r=2cm.
Now the area of circle is given as $a = \pi {r^2}$……………………… (4)
Putting the value in equation (3) we get
$a = \pi {(2)^2} = 4\pi c{m^2}$…………………. (5)
Clearly,
Required highlighted area = Area of square – Area of circle………………… (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.