A geometry problem by Mohammad Khaza

An alternative way to approach the question would be to use the fact that sides are directly proportional to each other, while areas are proportional to their squares. In other words, for two shapes $A$ and $B$ , the ratio of their sides is $A:B$ ; while the ratio of their areas is $A^2 : B^2$ .

Let the smaller circle be $A$ , and the larger circle be $B$ . Since $B$ 's radius is $2$ times of $A$ 's, $B = 2A$ . Using the rule described earlier, $B^2 = (2^2)(A^2)$ . The factor of $2^2$ tells us that $B$ is $4$ times larger than $A$ .

suppose, radius of the circle is =r and and area is = $πr^2$

so, new radius=2r and new area = $π(2r)^2$ = $4πr^2$ =4 x old area

Say the radius of the small original circle is $r$ , then the radius of the bigger circle is $2r$ . Now the area of the original circle is $\pi r^2$ and the area of the bigger circle is $\pi (2r)^2=4\pi r^2$ . That means that the area of the bigger circle is $\color{#D61F06}\large \boxed{4}$ times the area of the original circle.