A chord of a circle of radius subtends an angle of at the centre. Find the area of the corresponding segment of the circle. (Use and )

Understand the Question The area of a segment of a circle is found by subtracting the area of the corresponding triangle from the area of the sector: Given: Radius of the circle, Central angle, Step 1 · Find the Area of Triangle OST Draw perpendicular from centre to chord , which bisects both chord and . In right , with hypotenuse Length of chord Area of Step 2 · Find the Area of Sector OSUT Using the sector area formula with and Step 3 · Calculate the Area of the Segment Subtracting the area of from the area of sector Answer > Common Mistakes Approximation Value Error: Using instead of the specified or using instead of . Triangle Area Calculation Error: Forgetting that perpendicular from centre bisects chord , so the full base is .

Areas Related to Circles | Exercise 11.1

Question 7

A chord of a circle of radius $12\text{ cm}$ subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.

(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$ )

Solution

Understand the Question

The area of a segment of a circle is found by subtracting the area of the corresponding triangle from the area of the sector: $\text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle}$
Given:
- Radius of the circle, $r = 12\text{ cm}$
- Central angle, $\theta = 120^\circ$
- $\pi = 3.14$
- $\sqrt{3} = 1.73$

Step 1 · Find the Area of Triangle OST

Draw perpendicular $OV \perp ST$ from centre $O$ to chord $ST$ , which bisects both chord $ST$ and $\angle SOT$ . $\angle SOV = \dfrac{120^\circ}{2} = 60^\circ$ Diagram 1

In right $\triangle OVS$ , with hypotenuse $OS = 12\text{ cm}$

\begin{aligned} OV &= OS \times \cos 60^\circ \\[0.6em] &= 12 \times \dfrac{1}{2} \\[0.6em] &= 6\text{ cm} \end{aligned}

\begin{aligned} SV &= OS \times \sin 60^\circ \\[0.6em] &= 12 \times \dfrac{\sqrt{3}}{2} \\[0.6em] &= 6\sqrt{3}\text{ cm} \end{aligned}

Length of chord $ST$

\begin{aligned} ST &= 2 \times SV \\[0.6em] &= 2 \times 6\sqrt{3} \\[0.6em] &= 12\sqrt{3}\text{ cm} \end{aligned}

Area of $\triangle OST$

\begin{aligned} \text{Area of } \triangle OST &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times ST \times OV \\[0.6em] &= \dfrac{1}{2} \times 12\sqrt{3} \times 6 \\[0.6em] &= 36\sqrt{3}\text{ cm}^2 \\[0.6em] &= 36 \times 1.73 \\[0.6em] &= 62.28\text{ cm}^2 \end{aligned}

Step 2 · Find the Area of Sector OSUT

Using the sector area formula with $\theta = 120^\circ$ and $r = 12\text{ cm}$

\begin{aligned} \text{Area of sector OSUT} &= \dfrac{\theta}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{120^\circ}{360^\circ} \times \pi (12)^2 \\[0.6em] &= \dfrac{1}{3} \times 3.14 \times 144 \\[0.6em] &= 3.14 \times 48 \\[0.6em] &= 150.72\text{ cm}^2 \end{aligned}

Step 3 · Calculate the Area of the Segment

Subtracting the area of $\triangle OST$ from the area of sector $OSUT$

\begin{aligned} \text{Area of segment} &= \text{Area of sector OSUT} - \text{Area of } \triangle OST \\[0.6em] &= 150.72 - 62.28 \\[0.6em] &= 88.44\text{ cm}^2 \end{aligned}

Answer

$88.44\text{ cm}^2$

Common Mistakes

Approximation Value Error: Using $\pi = \dfrac{22}{7}$ instead of the specified $\pi = 3.14$ or using $\sqrt{3} \approx 1.732$ instead of $1.73$ .
Triangle Area Calculation Error: Forgetting that perpendicular from centre bisects chord $ST$ , so the full base is $ST = 2 \times SV = 12\sqrt{3}\text{ cm}$ .

Question 7

More questions in Exercise 11.1