Areas Related to Circles | Exercise 11.1

Question 4

A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding :

(i) minor segment

(ii) major sector. (Use $\pi = 3.14$ )

Solution

Understand the Question

A circle has radius $r = 10\text{ cm}$ and the chord subtends a right angle ( $\theta = 90^\circ$ ) at the centre $\text{O}$ .
Area of Minor Segment: Calculated by subtracting the area of the right-angled triangle $\text{OAB}$ from the area of the minor sector $\text{OAB}$ .
Area of Major Sector: Calculated by subtracting the area of the minor sector from the total area of the circle (or using central angle $360^\circ - 90^\circ = 270^\circ$ ).
Take $\pi = 3.14$ as specified in the problem.

(i) Find the area of the corresponding minor segment.

Step 1 · Calculate Area of Minor Segment

Given radius $r = 10\text{ cm}$ and central angle $\theta = 90^\circ$ . Diagram 1

Area of the minor sector

\begin{aligned} \text{Area of minor sector} &= \dfrac{\theta}{360^\circ} \times \pi r^2 \\[0.6em] &= \dfrac{90^\circ}{360^\circ} \times 3.14 \times (10\text{ cm})^2 \\[0.6em] &= \dfrac{1}{4} \times 3.14 \times 100\text{ cm}^2 \\[0.6em] &= 0.25 \times 314\text{ cm}^2 \\[0.6em] &= 78.5\text{ cm}^2 \end{aligned}

Since $\angle\text{AOB} = 90^\circ$ , $\triangle\text{OAB}$ is a right-angled triangle with $\text{base} = \text{height} = r = 10\text{ cm}$

\begin{aligned} \text{Area of } \triangle \text{OAB} &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times 10\text{ cm} \times 10\text{ cm} \\[0.6em] &= \dfrac{1}{2} \times 100\text{ cm}^2 \\[0.6em] &= 50\text{ cm}^2 \end{aligned}

Area of the minor segment

\begin{aligned} \text{Area of minor segment} &= \text{Area of minor sector} - \text{Area of } \triangle \text{OAB} \\[0.6em] &= 78.5\text{ cm}^2 - 50\text{ cm}^2 \\[0.6em] &= 28.5\text{ cm}^2 \end{aligned}

Answer

(i) $28.5\text{ cm}^2$

(ii) Find the area of the corresponding major sector. (Use $\pi = 3.14$ )

Step 1 · Calculate Area of Major Sector

Total area of the circle

\begin{aligned} \text{Area of circle} &= \pi r^2 \\[0.6em] &= 3.14 \times (10\text{ cm})^2 \\[0.6em] &= 3.14 \times 100\text{ cm}^2 \\[0.6em] &= 314\text{ cm}^2 \end{aligned}

Subtract the area of the minor sector

\begin{aligned} \text{Area of major sector} &= \text{Area of circle} - \text{Area of minor sector} \\[0.6em] &= 314\text{ cm}^2 - 78.5\text{ cm}^2 \\[0.6em] &= 235.5\text{ cm}^2 \end{aligned}

Answer

(ii) $235.5\text{ cm}^2$

Common Mistakes

Sector vs. Segment: Confusing the minor sector (the full slice enclosed by two radii and an arc) with the minor segment (the region between the chord and the arc).
Value of $\pi$ : Using $\pi = \dfrac{22}{7}$ instead of the explicitly instructed value $\pi = 3.14$ .
Segment vs. Major Sector: For part (ii), ensure you subtract the minor sector area from the total circle area, not the minor segment area.

More questions in Exercise 11.1