Measuring Space: Perimeter and Area | Exercise 6.3

Question 5

A chord of a circle of radius $15\text{ cm}$ subtends an angle of $60^\circ$ at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi \approx 3.14$ and $\sqrt{3} \approx 1.73$ .)

Solution

Understand the Question

Given a circle with radius $r = 15\text{ cm}$ and a chord subtending a central angle $\theta = 60^\circ$ .
The triangle formed by the center and the chord endpoints is equilateral because the two radii are equal and the included angle is $60^\circ$ .
Area of minor segment $= \text{Area of minor sector} - \text{Area of equilateral triangle}$ .
Area of major segment $= \text{Area of circle} - \text{Area of minor segment}$ .

Step 1 · Find the Area of the Minor Sector

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

$\text{Area of sector} = \dfrac{\theta}{360^\circ} \times \pi r^2$

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

Step 2 · Find the Area of the Triangle

Since the two radii are equal ( $OA = OB = 15\text{ cm}$ ) and the angle at the center is $60^\circ$ , $\Delta OAB$ is an equilateral triangle with side length $a = 15\text{ cm}$ .

$\text{Area of equilateral triangle} = \dfrac{\sqrt{3}}{4} a^2$

\begin{aligned} \text{Area of } \Delta OAB &= \dfrac{1.73}{4} \times 15^2 \\[0.6em] &= \dfrac{1.73}{4} \times 225 \\[0.6em] &= 97.3125\text{ cm}^2 \end{aligned}

Step 3 · Find the Area of the Minor Segment

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

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

Step 4 · Find the Area of the Circle

$\text{Area of circle} = \pi r^2$

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

Step 5 · Find the Area of the Major Segment

$\text{Area of major segment} = \text{Area of circle} - \text{Area of minor segment}$

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

Answer

$\text{Area of minor segment} = 20.4375\text{ cm}^2 \approx 20.44\text{ cm}^2$ , $\text{Area of major segment} = 686.0625\text{ cm}^2 \approx 686.06\text{ cm}^2$

Common Mistakes

Segment vs. Sector: Confusing a circular segment with a sector. A segment is formed by a chord and an arc, requiring subtraction of the triangle's area from the sector's area.
Triangle Type Recognition: Missing that an isosceles triangle with an apex angle of $60^\circ$ is equilateral, where area can be directly evaluated using $\dfrac{\sqrt{3}}{4} a^2$ .
Major Segment Calculation: Overcomplicating the major segment calculation instead of using $\text{Area of circle} - \text{Area of minor segment}$ .

Question 5

More questions in Exercise 6.3