A chord of a circle of radius subtends an angle of at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use and )

Understand the Question Given a circle with radius and a chord subtending an angle at the centre. Area of minor segment . Since the two radii are equal and the included angle is , is an equilateral triangle with side , whose area is . Area of major segment . Use the given values and . Step 1 · Calculate Area of Sector Given radius and angle . Step 2 · Calculate Area of Minor Segment In , (radii) and . Since , the base angles are equal: . Therefore, is an equilateral triangle. Now, calculate the area of the minor segment Step 3 · Calculate Area of Major Segment First, find the total area of the circle Now, calculate the area of the major segment Answer > Minor segment area Major segment area Common Mistakes Constant Substitution Error: Using or instead of the specified values and . Confusing Sector and Segment: Subtracting the triangle's area from the circle instead of from the corresponding sector when finding the minor segment. Equilateral Triangle Area Formula: Forgetting the factor of in the denominator of .

Areas Related to Circles | Exercise 11.1

Question 6

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.

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

Solution

Understand the Question

Given a circle with radius $r = 15\text{ cm}$ and a chord subtending an angle $\theta = 60^\circ$ at the centre.
Area of minor segment $= \text{Area of minor sector} - \text{Area of } \triangle \text{OPQ}$ .
Since the two radii are equal and the included angle is $60^\circ$ , $\triangle \text{OPQ}$ is an equilateral triangle with side $15\text{ cm}$ , whose area is $\dfrac{\sqrt{3}}{4} r^2$ .
Area of major segment $= \text{Area of circle} - \text{Area of minor segment}$ .
Use the given values $\pi = 3.14$ and $\sqrt{3} = 1.73$ .

Step 1 · Calculate Area of Sector

Given radius $r = 15\text{ cm}$ and angle $\theta = 60^\circ$ . Diagram 1

\begin{aligned} \text{Area of sector} &= \dfrac{\theta}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{60^\circ}{360^\circ} \times 3.14 \times (15)^2 \\[0.6em] &= \dfrac{1}{6} \times 3.14 \times 225 \\[0.6em] &= \dfrac{706.5}{6} \\[0.6em] &= 117.75\text{ cm}^2 \end{aligned}

Step 2 · Calculate Area of Minor Segment

In $\triangle \text{OPQ}$ , $\text{OP} = \text{OQ} = 15\text{ cm}$ (radii) and $\angle \text{POQ} = 60^\circ$ . Since $\text{OP} = \text{OQ}$ , the base angles are equal: $\angle \text{OPQ} = \angle \text{OQP} = 60^\circ$ . Therefore, $\triangle \text{OPQ}$ is an equilateral triangle.

\begin{aligned} \text{Area of } \triangle \text{OPQ} &= \dfrac{\sqrt{3}}{4} \times (\text{side})^2 \\[0.6em] &= \dfrac{1.73}{4} \times (15)^2 \\[0.6em] &= \dfrac{1.73 \times 225}{4} \\[0.6em] &= \dfrac{389.25}{4} \\[0.6em] &= 97.3125\text{ cm}^2 \end{aligned}

Now, calculate the area of the minor segment

\begin{aligned} \text{Area of minor segment} &= \text{Area of sector} - \text{Area of } \triangle \text{OPQ} \\[0.6em] &= 117.75 - 97.3125 \\[0.6em] &= 20.4375\text{ cm}^2 \end{aligned}

Step 3 · Calculate Area of Major Segment

First, find the total area of the circle

\begin{aligned} \text{Area of circle} &= \pi r^2 \\[0.6em] &= 3.14 \times (15)^2 \\[0.6em] &= 3.14 \times 225 \\[0.6em] &= 706.5\text{ cm}^2 \end{aligned}

Now, calculate the area of the major segment

\begin{aligned} \text{Area of major segment} &= \text{Area of circle} - \text{Area of minor segment} \\[0.6em] &= 706.5 - 20.4375 \\[0.6em] &= 686.0625\text{ cm}^2 \end{aligned}

Answer

Minor segment area $= 20.4375\text{ cm}^2$

Major segment area $= 686.0625\text{ cm}^2$

Common Mistakes

Constant Substitution Error: Using $\pi = \dfrac{22}{7}$ or $\sqrt{3} = 1.732$ instead of the specified values $\pi = 3.14$ and $\sqrt{3} = 1.73$ .
Confusing Sector and Segment: Subtracting the triangle's area from the circle instead of from the corresponding sector when finding the minor segment.
Equilateral Triangle Area Formula: Forgetting the factor of $4$ in the denominator of $\dfrac{\sqrt{3}}{4} a^2$ .

Question 6

More questions in Exercise 11.1