Question

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is $1256c{m}^2$.

Answer 1

Rhombus is a cyclic Quadrilateral here

$\Rightarrow$ Opposite angles sum$={180}^o\Rightarrow A+C={180}^o$

Also as $AB||CD\Rightarrow A+B={180}^o\Rightarrow B=C$

$\Rightarrow$ Adjacent angles of rohmbus are equal $\Rightarrow$ it is square

$\angle B={90}^o\Rightarrow \angle B$ is a angle in a semicircle.

$\Rightarrow$ AC and BD are diameters.

$\Rightarrow$ Diagonals bisect at o, center of circle.

In $\Delta AOB,A{B}^2=A{O}^2+O{B}^2=r^2+r^2=2r^2\Rightarrow AB\sqrt{2}r$

Area of circle $=\pi r^2=1256\Rightarrow r^2=\frac{1256}{r}$

Area of square $=A{B}^2=2r^2=\frac{2\times 1256}{\pi }=\frac{2\times 1256\times 7}{22}=799.27c{m}^2$