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

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(C) Let the radius of the circle be $r$.Given that,Area of the circle $= 1256 \, cm^{2}$.$\pi r^{2} = 1256$$r^{2} = \frac{1256}{3.14} = 400$$r = \sqrt

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(C) Let the radius of the circle be $r$.Given that,Area of the circle $= 1256 \, cm^{2}$.$\pi r^{2} = 1256$$r^{2} = \frac{1256}{3.14} = 400$$r = \sqrt{400} = 20 \, cm$.Since all the vertices of a rhombus lie on a circle,the rhombus must be a square because its diagonals are diameters of the circle and they bisect each other at $90^{\circ}$.Thus,the diagonals $d_{1}$ and $d_{2}$ are equal to the diameter of the circle.$d_{1} = d_{2} = 2 	imes r = 2 	imes 20 = 40 \, cm$.Area of the rhombus $= \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 the rhombus is $800 \, cm^{2}$.

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

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