The area of a circle is increased by 22, cm^{ | vedclass

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(A) Let the original radius of the circle be $r\, cm$.The area of the original circle is $A_1 = \pi r^{2}$.When the radius is increased by $1\, cm$,th

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$3$

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(A) Let the original radius of the circle be $r\, cm$.The area of the original circle is $A_1 = \pi r^{2}$.When the radius is increased by $1\, cm$,the new radius becomes $(r + 1)\, cm$.The area of the new circle is $A_2 = \pi(r + 1)^{2}$.According to the problem,the increase in area is $22\, cm^{2}$,so $A_2 - A_1 = 22$.$\pi(r + 1)^{2} - \pi r^{2} = 22$$\pi(r^{2} + 2r + 1 - r^{2}) = 22$$\pi(2r + 1) = 22$Substituting $\pi = \frac{22}{7}$,we get:$\frac{22}{7}(2r + 1) = 22$Dividing both sides by $22$:$\frac{1}{7}(2r + 1) = 1$$2r + 1 = 7$$2r = 6$$r = 3\, cm$.Thus,the original radius of the circle is $3\, cm$.

The area of a circle is increased by $22\, cm^{2}$ when its radius is increased by $1\, cm$. The original radius of the circle is.......$cm$.

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