In $Fig.$ a square of diagonal $8\, cm$ is inscribed in a circle. Find the area of the shaded region.
Let the side of a square be a and the radius of circle be $r$.
Given that, length of diagonal of square $=8\, cm$
$\Rightarrow$ $a \sqrt{2}=8$
$\Rightarrow$ $a=4 \sqrt{2} \, cm$
Now, Diagonal of a square $=$ Diameter of a circle
$\Rightarrow$ Diameter of circle $=8$
$\Rightarrow$ Radius of circle $=r=\frac{\text { Diameter }}{2}$
$\Rightarrow$ $r=\frac{8}{2}=4 \,cm$
$\therefore$ Area of circle $=\pi r^{2}=\pi(4)^{2}$
$=16 \pi \times\, cm ^{2}$
and Area of square $=a^{2}=(4 \sqrt{2})^{2}$
$=32 cm ^{2}$
So, the area of the shaded region $=$ Area of circle $-$ Area of square
$=(16 \pi-32) \, cm ^{2}$
Hence, the required area of the shaded region is $(16 \pi-32) \, cm ^{2}$.
In $Fig.$ arcs have been drawn with radii $14\, cm$ each and with centres $P , Q$ and $R$. Find the area of the shaded region. (in $cm ^{2}$)
The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?
Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii $15 \,cm$ and $18 \,cm$ (in $cm$)
Floor of a room is of dimensions $5 \,m \times 4 \,m$ and it is covered with circular tiles of diameters $50 \,cm$ each as shown in $Fig.$ Find the area of floor that remains uncovered with tiles. (Use $\pi=3.14)$ (in $m ^{2}$)
The maximum area of a triangle inscribed in a semicircle having radius $10\,cm$ is $\ldots \ldots \ldots . . cm ^{2} .$