All the vertices of a rhombus lie on a circle | vedclass

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Question

Let the radius of the circle be $r.$

Given that, Area of the circle $=1256\, cm ^{2}$

$\pi r^{2}=1256$

$r^{2}=\frac{1256}{\pi}=\frac{1256}{31

Vedclass · Accepted Answer

$800$

Vedclass · Answer

Let the radius of the circle be $r.$

Given that, Area of the circle $=1256\, cm ^{2}$

$\pi r^{2}=1256$

$r^{2}=\frac{1256}{\pi}=\frac{1256}{314}=400$

$r^{2}=(20)^{2}$

$r=20\, cm$

$	herefore$ So, the radius of circle is $20\, cm$

$\Rightarrow$ Diameter of circle $=2 	imes$ Radius

$=2 	imes 20$

$=40 \,cm$

since, all the vertices of a rhombus lie on a circle that means each diagonal of a rhombus must pass through the centre of a circle that is why both diagonals are equal and same as the diameter of the given circle.

Let $d,$ and $d_{2}$ be the diagonals of the rhombus.

$	herefore \quad d_{1}=d_{2}=$ Diameter of circle $=40 \,cm$

So, $\quad$ Area of mombus $=\frac{1}{2} 	imes d_{1} 	imes d_{2}$

$=\frac{1}{2} 	imes 40 	imes 40$

$=20 	imes 40=800 \,cm ^{2}$

Hence, the required area of rhombus is $800\, cm ^{2}$

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is $1256 \,cm ^{2}$. (Use $\pi=3.14$ ). (in $cm ^{2}$)

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