Class 10 NCERT Solutions
Chapter 11: Areas Related to Circles
Master the geometry of sectors and segments, the derivation of arc lengths, and the logic of circular regions with our step-by-step logic.
Exercise 11.1
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60°$.
$$\text{Area of sector} = \frac{\theta}{360°}\times\pi r^2 = \frac{60°}{360°}\times\frac{22}{7}\times36 = \frac{132}{7} \text{ cm}^2$$
2. Find the area of a quadrant of a circle whose circumference is 22 cm.
A quadrant corresponds to $\theta = 90°$. First find the radius from the circumference:
$$2\pi r = 22 \Rightarrow r = \frac{22}{2\pi} = \frac{7}{2} \text{ cm}$$ $$\text{Area of quadrant} = \frac{1}{4}\pi r^2 = \frac{1}{4}\times\frac{22}{7}\times\frac{49}{4} = \frac{77}{8} \text{ cm}^2$$
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
In 5 minutes the minute hand rotates through $\dfrac{360°}{60} \times 5 = 30°$.
$$\text{Area covered} = \frac{30°}{360°}\times\frac{22}{7}\times14^2 = \frac{1}{12}\times22\times2\times14 = \frac{154}{3} \text{ cm}^2$$
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use $\pi = 3.14$)
Area of minor sector ($\theta = 90°$, $r = 10$ cm):
$$= \frac{90}{360}\times3.14\times100 = \frac{314}{4} = 78.5 \text{ cm}^2$$(i) Area of minor segment:
$$= 78.5 – \frac{1}{2}\times10\times10 = 78.5 – 50 = 28.5 \text{ cm}^2$$(ii) Area of major sector:
$$= \pi r^2 – 78.5 = 314 – 78.5 = 235.5 \text{ cm}^2$$
5. In a circle of radius 21 cm, an arc subtends an angle of $60°$ at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.
(i) Length of arc $= \dfrac{\theta\pi r}{180°} = \dfrac{60°\times\pi\times21}{180°} = 7\pi = 22$ cm
(ii) Area of sector $= \dfrac{60°}{360°}\times\dfrac{22}{7}\times21^2 = \dfrac{1}{6}\times22\times3\times21 = 231$ cm²
(iii) Area of segment $= \text{Area of sector} – \dfrac{1}{2}r^2\sin 60°$
$$= 231 – \frac{1}{2}\times441\times\frac{\sqrt{3}}{2} = \left(231-\frac{441\sqrt{3}}{4}\right) \text{ cm}^2$$
6. A chord of a circle of radius 15 cm subtends an angle of $60°$ at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Since $\angle\text{AOB} = 60°$ and OA = OB (radii), triangle OAB is equilateral.
Area of minor segment:
$$= \frac{60°}{360°}\times3.14\times225 – \frac{1}{2}\times225\times\sin 60° = 117.75 – 97.3125 = 20.4375 \text{ cm}^2$$Area of major segment:
$$= 3.14\times225 – 20.4375 = 706.5 – 20.4375 = 686.0625 \text{ cm}^2$$
7. A chord of a circle of radius 12 cm subtends an angle of $120°$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)
Drop perpendicular OL from O to chord AB. Then $\angle\text{LOB} = 60°$.
$$\text{OL} = 12\cos 60° = 6 \text{ cm}, \quad \text{LB} = 12\sin 60° = 6\sqrt{3} \text{ cm}$$ $$\text{AB} = 2\text{LB} = 12\sqrt{3} \text{ cm}$$ $$\text{Area of segment} = \frac{120°}{360°}\times3.14\times144 – \frac{1}{2}\times12\sqrt{3}\times6$$ $$= [3.14\times48 – 36\times1.73] = [150.72-62.28] = 88.44 \text{ cm}^2$$
8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find: (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use $\pi = 3.14$)
The horse can graze a quarter-circle (the corner angle of the square is $90°$).
(i)
$$\text{Area} = \frac{90°}{360°}\times3.14\times5^2 = \frac{1}{4}\times3.14\times25 = 19.625 \text{ m}^2$$(ii)
$$\text{Increase} = \frac{1}{4}\pi(10^2-5^2) = \frac{3.14}{4}\times75 = 58.875 \text{ m}^2$$
9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
(i)
$$\text{Total length} = \text{circumference}+5\times\text{diameter} = \pi\times35+5\times35 = 110+175 = 285 \text{ mm}$$(ii) Each sector has central angle $\dfrac{360°}{10} = 36°$.
$$\text{Area of each sector} = \frac{36°}{360°}\times\frac{22}{7}\times\left(\frac{35}{2}\right)^2 = \frac{385}{4} \text{ mm}^2$$
10. An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Each pair of consecutive ribs forms a sector with central angle $\dfrac{360°}{8} = 45°$.
$$\text{Area between two ribs} = \frac{45°}{360°}\times\frac{22}{7}\times45^2 = \frac{22275}{28} \text{ cm}^2$$
11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of $115°$. Find the total area cleaned at each sweep of the blades.
$$\text{Total area} = 2\times\frac{115°}{360°}\times\frac{22}{7}\times25^2 = \frac{23\times11\times25\times25}{18\times7} = \frac{158125}{126} \text{ cm}^2$$
12. To warn ships for underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle $80°$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)
$$\text{Area of sea} = \frac{80°}{360°}\times3.14\times(16.5)^2 = \frac{2}{9}\times3.14\times272.25 \approx 189.97 \text{ km}^2$$
13. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use $\sqrt{3} = 1.7$)
Each design is a segment with central angle $60°$. Area of 6 segments:
$$= 6\times\left[\frac{60°}{360°}\times\frac{22}{7}\times28^2 – \frac{1}{2}\times28^2\times\sin 60°\right]$$ $$= \left[\frac{22}{7}\times784 – \frac{3\sqrt{3}}{2}\times784\right] = (2464-1999.2) = 464.8 \text{ cm}^2$$ $$\text{Cost} = ₹(464.8\times0.35) = ₹\,162.68$$
14. Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is:
(A) $\frac{p}{180}\times 2\pi R$ (B) $\frac{p}{180}\times\pi R^2$ (C) $\frac{p}{360}\times 2\pi R$ (D) $\frac{p}{720}\times 2\pi R^2$
$$\text{Area of sector} = \frac{p}{360}\times\pi R^2 = \frac{p}{720}\times 2\pi R^2$$
Answer: (D)
