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

Hint:Vertices of cyclic quadrilateral lie on a circle and for cyclic quadrilateral, sum of opposite angles is equal to ${180^0}$. Apply the properties of rhombus and find the relation between radius of circle and area of rhombus.

Complete step-by-step answer:
Since it is given rhombus is a cyclic quadrilateral.
Let ${d_1}{\text{ and }}{d_2}$be diagonals of rhombus
Sum of opposite angles = ${180^0}$
$\angle A + \angle C = {180^0}$
Also AB || CD
$
   \Rightarrow \angle A + \angle B = {180^0} \\
   \Rightarrow \angle B = \angle C \\
$
Now, as adjacent angles are equal, it is a square.
$\angle B = {90^0}$
$\angle B$is angle in semicircle
AC and BD are diameter of circle.
Now, Area of circle = 1256
$
  \pi {r^2} = 1256 \\
  {r^2} = \dfrac{{1256}}{{3.14}} \\
  {r^2} = 400 \\
  r = \sqrt {400} \\
  r = 20cm \\
$
Diameter of circle = 2r = 40cm $ \Rightarrow {d_1} = {d_2} = 40cm$
$\therefore $Area of rhombus
$
   = \dfrac{1}{2} \times {d_1} \times {d_2} \\
   = \dfrac{1}{2} \times 40 \times 40 \\
   = 800c{m^2} \\
    \\
$

Note: A cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic. Students must remember the formula for the area of some common geometrical figure such as circle and rhombus.