Inner-circle 2902

The student should remove the inner circle with a radius of 3 cm from the circle with a radius of 7 cm. How much of the area of the large circle will be removed? Express result in percent.

Correct answer:

p =  0.1837 %

Step-by-step explanation:

$$\begin{gathered} r_1=7\text{cm} \\ {r}_2=3\text{cm} \\ A=\pi \cdot r_1^2=3.1416\cdot 7^2\doteq 153.938{\text{cm}}^2 \\ B=\pi \cdot r_2^2=3.1416\cdot 3^2\doteq 28.2743{\text{cm}}^2 \\ x=B/A=28.2743/153.938=\frac{9}{49}\doteq 0.1837 \\ p=100\cdot x=100\cdot 0.1837=\frac{900}{49}\doteq 18.3673=0.1837\% \end{gathered}$$

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