Question

A chord of a circle, of radius 14 cm, subtends an angle of $60^\circ$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

Hint:

Whenever the central angle of a sector is $60^\circ$, the triangle formed by the center and the chord endpoints is always an equilateral triangle, so the chord length is always equal to the radius!

Answer 1

Step 1: Find the sector area directly using the simplified fraction $\frac{60^\circ}{360^\circ}=\frac16$.
\[ \text{Area}_{\text{sector}} = \frac16 \times \frac{22}{7} \times 14^2 = \frac16 \times 22 \times 28 = \frac{308}{3}\text{ cm}^2 \]
Step 2: Recognise the triangle formed by the two radii and the chord as equilateral, since the central angle is exactly $60^\circ$ with two equal sides.
This immediately gives chord length $AB = r = 14\text{ cm}$, without needing to find the base angles separately.
Step 3: Find the arc length using the same $\frac16$ fraction.
\[ l = \frac16 \times 2 \times \frac{22}{7} \times 14 = \frac16 \times 88 = \frac{44}{3}\text{ cm} \]
Step 4: Add the arc length and chord length for the segment perimeter, and conclude.
\[ \text{Perimeter} = \frac{44}{3}+14 = \frac{44+42}{3} = \frac{86}{3}\text{ cm} \]
\[ \boxed{\text{Sector area} = \frac{308}{3}\text{ cm}^2,\ \text{Segment perimeter} = \frac{86}{3}\text{ cm}} \]