Find the difference of the areas of a sector of angle \(120^\circ\) and its corresponding major sector of a circle of radius 21 cm.
\(147\pi\,\text{cm}^2 \;\approx\; 461.8\,\text{cm}^2\)
Step 1: Recall the formula for the area of a sector of a circle:
\[ A = \dfrac{\theta}{360^\circ} \times \pi r^2 \]
where:
Step 2: In this problem, the radius \(r = 21\,\text{cm}\).
Step 3: The given minor sector has angle \(120^\circ\).
So its corresponding major sector will have angle:
\[ 360^\circ - 120^\circ = 240^\circ \]
Step 4: Write the difference of the two areas:
\[ \text{Difference} = A_{240^\circ} - A_{120^\circ} \]
Step 5: Apply the formula:
\[ A_{240^\circ} - A_{120^\circ} = \dfrac{240}{360}\pi r^2 - \dfrac{120}{360}\pi r^2 \]
Step 6: Simplify the fractions:
\[ = \dfrac{240 - 120}{360}\pi r^2 = \dfrac{120}{360}\pi r^2 \]
Step 7: Substitute \(r = 21\,\text{cm}\):
\[ = \dfrac{120}{360} \times \pi \times 21^2 \]
Step 8: Simplify step by step:
So,
\[ = \tfrac{1}{3} \times \pi \times 441 = 147\pi\,\text{cm}^2 \]
Step 9: Approximate using \(\pi \approx 3.1416\):
\[ 147 \times 3.1416 \approx 461.8\,\text{cm}^2 \]
Final Answer: The difference of the areas is \(147\pi\,\text{cm}^2 \;\approx\; 461.8\,\text{cm}^2\).