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: Understanding the Question:
We are given a circle of radius $r = 14\text{ cm}$.
A chord of this circle subtends an angle of $\theta = 60^\circ$ at the center.
We need to find the area of the smaller sector and the perimeter of the smaller segment formed by this chord.

Step 2: Key Formula or Approach:
- Area of the smaller sector is given by:
\[ \text{Area}_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \]
- The perimeter of the smaller segment consists of the curved arc length ($l$) of the sector plus the straight length of the chord ($AB$):
\[ \text{Perimeter}_{\text{segment}} = \text{Arc length } (l) + \text{Chord length } (AB) \]
where:
\[ l = \frac{\theta}{360^\circ} \times 2\pi r \]

Step 3: Detailed Explanation:

• Calculate the Area of the smaller sector:
\[ \text{Area}_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 \]
\[ \text{Area}_{\text{sector}} = \frac{1}{6} \times \frac{22}{7} \times 196 \]
\[ \text{Area}_{\text{sector}} = \frac{1}{6} \times 22 \times 28 = \frac{308}{3}\text{ cm}^2 \approx 102.67\text{ cm}^2 \]

• Calculate the Arc length ($l$):
\[ l = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 \]
\[ l = \frac{1}{6} \times 2 \times 22 \times 2 = \frac{44}{3}\text{ cm} \approx 14.67\text{ cm} \]

• Determine the Chord length ($AB$):
Consider triangle $\Delta OAB$, where $OA = OB = r = 14\text{ cm}$ and $\angle AOB = 60^\circ$.
Since $OA = OB$, the angles opposite to them are equal:
\[ \angle OAB = \angle OBA \]
The sum of angles in $\Delta OAB$ is $180^\circ$:
\[ \angle OAB + \angle OBA + 60^\circ = 180^\circ \implies 2\angle OAB = 120^\circ \implies \angle OAB = 60^\circ \]
Since all three angles are $60^\circ$, $\Delta OAB$ is an equilateral triangle.
Therefore:
\[ AB = OA = OB = 14\text{ cm} \]

• Calculate the perimeter of the smaller segment:
\[ \text{Perimeter} = l + AB \]
\[ \text{Perimeter} = \frac{44}{3} + 14 = \frac{44 + 42}{3} = \frac{86}{3}\text{ cm} \approx 28.67\text{ cm} \]


Step 4: Final Answer:
The area of the smaller sector is $\frac{308}{3}\text{ cm}^2$ (or $102.67\text{ cm}^2$) and the perimeter of the smaller segment is $\frac{86}{3}\text{ cm}$ (or $28.67\text{ cm}$).