Answer:
Given that the area of the circle is \[1256\text{ }c{{m}^{2}}\]. \[\because \] Area of the circle \[=\pi {{r}^{2}}\] \[1256=\frac{3.14}{100}\times {{r}^{2}}\] \[{{r}^{2}}=\frac{1256\times 100}{314}\] \[r=\sqrt{400}\] \[r=20\,\,cm\] Now, ABCD are the vertices of a rhombus. \[\therefore \angle A=\angle C\] ...(i) [opposite angles of rhombus] But ABCD lie on the circle. So, ABCD is called cyclic quadrilateral \[\therefore \angle A+\angle C=180{}^\circ \] ...(ii) On using equation (i), we get \[\angle A+\angle A=180{}^\circ \] \[2\angle A=180{}^\circ \] \[\angle A=90{}^\circ \] so, \[\angle C=90{}^\circ \] [From eq. (i)] \[\therefore \text{ }ABCD\] is square. So, BD is a diameter of circle. [\[\because \] The angle in a semicircle is a right angle triangle] \[\text{Now, Area of rhombus =}\frac{1}{2}\times \text{ product of diagonals}\] \[\text{=}\frac{1}{2}\times 40\times 40\text{ }\] \[=800\,\,c{{m}^{2}}\] Hence, Area of rhombus is \[800\text{ }c{{m}^{2}}\].
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