Question 7

We will find the area of the sector and then subtract the area of the triangle.

Step 1 — Find the Area of the Sector

Let's call the center of the circle O. Let the chord be AB. The radius is r. The angle at the center is 60°. The area of a sector is a fraction of the circle's area.

$\text{Area of sector OAB} = \frac{\text{angle}}{360^\circ} \times \pi r^2$

$= \frac{60^\circ}{360^\circ} \times \pi r^2$

$= \frac{1}{6} \times \pi r^2$

$\boxed{\frac{\pi r^2}{6}}$

Step 2 — Find the Area of the Triangle

The triangle formed by the radii and the chord is $\triangle \text{OAB}$. Sides OA and OB are both radii, so they are equal to r. The angle $\angle \text{AOB}$ is 60°. Since two sides are equal, $\triangle \text{OAB}$ is an isosceles triangle. The base angles $\angle \text{OAB}$ and $\angle \text{OBA}$ are equal. The sum of angles in a triangle is 180°. So, $\angle \text{OAB} = \angle \text{OBA} = (180^\circ - 60^\circ) / 2 = 60^\circ$. All three angles are 60°. This means $\triangle \text{OAB}$ is an equilateral triangle. All sides are equal to r.

$\text{Area of } \triangle \text{OAB} = \frac{\sqrt{3}}{4} \times (\text{side})^2$

$= \frac{\sqrt{3}}{4} r^2$

$\boxed{\frac{\sqrt{3} r^2}{4}}$

Step 3 — Find the Area of the Minor Segment

The area of the minor segment is the area of the sector minus the area of the triangle.

$\text{Area of minor segment} = \text{Area of sector OAB} - \text{Area of } \triangle \text{OAB}$

$= \frac{\pi r^2}{6} - \frac{\sqrt{3} r^2}{4}$

We can factor out r².

$= r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)$

$\boxed{r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)}$

Answer

(i) The area of the corresponding minor segment of the circle is $r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)$.