Question

The difference between the area of outer and the inner square of a circle is 63cm^2. And "O" is the centre then find the area of the circle

Answer 1

The area of circle=99 square cm

Step-by-step explanation:

Let r be the radius of circle

Diagonals of square are  perpendicular bisect to each other

Side of inner square=$\sqrt{r^2+r^2}=r\sqrt 2$

By using Pythagoras theorem

$(hypotenuse)^2=(base)^2+(perpendicular\;side)^2$

Area of square=$side\times side$

Area of inner square =$r\sqrt 2\times r\sqrt 2=2r^2$

Side of outer square=Diameter of circle=2r

Area of outer square=$(2r)^2=4r^2$

Difference between the area of outer square and area of inner square=63 square cm

$4r^2-2r^2=63$

$2r^2=63$

$r^2=\frac{63}{2}$

Area of circle=$\pi r^2$

Where $\pi=\frac{22}{7}$

Substitute the values then we get

Area of circle=$\frac{22}{7}\times \frac{63}{2}=99cm^2$

Area of circle=99 square cm

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