Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of \(60^\circ\) (Use \(\pi=3.14\)).
\(\approx 13.01\,\text{cm}^2\)
Step 1: Write the formula for the area of a sector.
\(A_{sector} = \dfrac{\theta}{360^\circ} \times \pi r^2\)
Here, radius \(r = 12\,\text{cm}\), angle \(\theta = 60^\circ\), and \(\pi = 3.14\).
\(A_{sector} = \dfrac{60}{360} \times 3.14 \times (12)^2\)
\(= \dfrac{1}{6} \times 3.14 \times 144\)
\(= 75.36\,\text{cm}^2\)
Step 2: Write the formula for the area of an isosceles triangle inside the sector.
That triangle is formed by two radii (each 12 cm) and the included angle \(60^\circ\).
Formula: \(A_{triangle} = \dfrac{1}{2} r^2 \sin(\theta)\)
\(= \dfrac{1}{2} (12)^2 \sin 60^\circ\)
\(= \dfrac{1}{2} \times 144 \times \dfrac{\sqrt{3}}{2}\)
\(= 36\sqrt{3}\,\text{cm}^2\)
Now approximate \(\sqrt{3} \approx 1.732\).
So, \(A_{triangle} = 36 \times 1.732 \approx 62.35\,\text{cm}^2\).
Step 3: Find the area of the segment.
\(A_{segment} = A_{sector} - A_{triangle}\)
\(= 75.36 - 62.35\)
\(\approx 13.01\,\text{cm}^2\)
Final Answer: Area of the segment = \(13.01\,\text{cm}^2\)