all the vertices of a rhombus lie on a circle. find the area of the rhombus if the area of the circle is 1256cm^{2}

**Given: **All the vertices of a rhombus lie on the circle

Let ABCD be the rhuombus.

In a rhombus opposite angles are equal

⇒ ∠A = ∠C ...... (1)

but in a cyclic quadrilateral opposite angles are supplementary

⇒ ∠A + ∠C = 180° ...... (2)

from (1) and (2) we get

Hence ABCD is a square.

and diagonal AC is the diameter (∵ converse of angle in a semicircle is right angle)

Now area of circle = 1256 cm^{2}

⇒ π radius^{2} = 1256 cm^{2}

Now in right ∆ ABC

AB^{2} + BC^{2} = AC^{2} = (Diameter)^{2}

⇒ 2(side)^{2} = (2 radius)^{2}

⇒ side^{2} = 2 radius^{2}

Hence area of ABCD = side^{2} = 2 radius^{2}

**
**