question_answer

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if the area of the circle is\[1256\text{ }c{{m}^{2}}\]. [Use\[\pi =3.14\]]

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 \[800\text{ }c{{m}^{2}}\].