Question

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

Answer 1

Given that the area of the circle is 1256 $c{m}^2$.

$\therefore$ 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.

ASo,ABCD is called cyclic quadrilateral

$\therefore$ $\angle A+\angle C={180}^{\circ }$ ....(ii)

On using equation (i), we get

$\angle A+\angle A={180}^{\circ }$

$2\angle A={180}^{\circ }$

$\angle A={90}^{\circ }$

$\angle C={90}^{\circ }$ [From eq.(i)]

so,

$\therefore$ ABCD is square.

So,BD is a diameter of circle.

$[\because$ The angle in a semicircle is a right angle triangle]

Now,Area of rhombus=$\frac{1}{2}\times \text{Product of diagonals}$

=$\frac{1}{2}\times 40\times 40$

=$800c{m}^2$

Hence,Area of rhombus is $800c{m}^2$