Measuring Space: Perimeter and Area | Exercise 6.3

Question 7

A chord of a circle of radius $r$ subtends an angle of $60^\circ$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $\pi r^2 \left( \dfrac{1}{6} - \dfrac{\sqrt{3}}{4} \right)$ .

Solution

Understand the Question

The area of a minor segment is obtained by subtracting the area of the corresponding triangle from the area of the sector: $\text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle}$
Given a circle of radius $r$ $r$ and a central angle $\theta = 60^\circ$ $θ = 6 0^{\circ}$ :
- $\text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2$
- The triangle formed by the two radii and the chord is equilateral (since the central angle is $60^\circ$ and the two sides are equal radii $r$ ), so $\text{Area of triangle} = \dfrac{\sqrt{3}}{4}r^2$ .

Step 1 · Find Area of Sector

Let the center of the circle be $\text{O}$ and the chord be $\text{AB}$ . Radius is $r$ and central angle is $60^\circ$ . Diagram 1

\begin{aligned} \text{Area of sector OAB} &= \dfrac{\text{angle}}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{60^\circ}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{\pi r^2}{6} \end{aligned}

Step 2 · Find Area of Triangle

In $\triangle \text{OAB}$ , $\text{OA} = \text{OB} = r$ , so $\angle \text{OAB} = \angle \text{OBA}$ .

$\angle \text{OAB} = \angle \text{OBA} = \dfrac{180^\circ - 60^\circ}{2} = 60^\circ$

Since all angles are $60^\circ$ , $\triangle \text{OAB}$ is an equilateral triangle of side $r$ .

\begin{aligned} \text{Area of } \triangle \text{OAB} &= \dfrac{\sqrt{3}}{4} \times (\text{side})^2 \\[0.6em] &= \dfrac{\sqrt{3}}{4} r^2 \end{aligned}

Step 3 · Find Area of Minor Segment

\begin{aligned} \text{Area of minor segment} &= \text{Area of sector OAB} - \text{Area of } \triangle \text{OAB} \\[0.6em] &= \dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}r^2}{4} \\[0.6em] &= r^2 \left( \dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4} \right) \end{aligned}

Answer

$\text{Area of minor segment} = r^2 \left( \dfrac{\pi}{6} - \dfrac{\sqrt{3}}{4} \right)$

Common Mistakes

Triangle Type Assumption: Assuming $\triangle \text{OAB}$ is a right-angled triangle instead of proving it is equilateral using $\text{OA} = \text{OB} = r$ and $\angle \text{AOB} = 60^\circ$ .
Factoring $\pi$ : Factoring out $\pi r^2$ incorrectly across both terms when $\pi$ is only present in the sector term, not in the equilateral triangle's area formula $\dfrac{\sqrt{3}}{4}r^2$ .

Question 7

More questions in Exercise 6.3