Problem: Circle I passes through the center of, and is tangent to, circle II. The area of circle III is 444 square inches. Then the area of circle II, in square inches, is:
Answer Choices:
A. 888
B. 828 \sqrt{2}82​
C. 8π8 \sqrt{\pi}8π​
D. 161616
E. 16216 \sqrt{2}162​ Solution:
Area II Area I=πR2πr2=(2r)2r2=4\dfrac{\text { Area II }}{\text { Area } I}=\dfrac{\pi \mathrm{R}^{2}}{\pi \mathrm{r}^{2}}=\dfrac{(2 r)^{2}}{r^{2}}=4 Area I Area II ​=πr2πR2​=r2(2r)2​=4
∴\therefore∴ Area II=4I I=4II=4 Area I=4⋅4=16I=4 \cdot 4=16I=4⋅4=16