NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles Exercise 11.1#
Exercise 11.1 covers areas of sectors, segments, and combinations of plane figures involving circles.
Key Formulas:
- Length of arc $= \dfrac{\theta}{360°} \times 2\pi r$
- Area of sector $= \dfrac{\theta}{360°} \times \pi r^2$
- Area of minor segment $=$ Area of sector $-$ Area of triangle
- Use $\pi = \dfrac{22}{7}$ unless stated otherwise.
1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°. #
Answer
∴ Area of sector $= \dfrac{132}{7}$ cm²
2. Find the area of a quadrant of a circle whose circumference is 22 cm. #
Answer
Circumference $= 2\pi r = 22$
$$r = \frac{22}{2\pi} = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} \text{ cm}$$Area of quadrant $= \dfrac{90°}{360°} \times \pi r^2 = \dfrac{1}{4} \times \dfrac{22}{7} \times \dfrac{49}{4}$
$$= \frac{22 \times 49}{4 \times 4 \times 7} = \frac{22 \times 7}{16} = \frac{154}{16} = \frac{77}{8} \text{ cm}^2$$∴ Area of quadrant $= \dfrac{77}{8}$ cm²
3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. #
Answer
In 60 minutes, minute hand completes 360°.
In 5 minutes, angle swept $= \dfrac{360°}{60} \times 5 = 30°$
$$\text{Area swept} = \frac{30}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{12} \times \frac{22}{7} \times 196$$$$= \frac{22 \times 196}{12 \times 7} = \frac{22 \times 28}{12} = \frac{616}{12} = \frac{154}{3} \text{ cm}^2$$∴ Area swept $= \dfrac{154}{3}$ cm²
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 π = 3.14) #
(i) Minor segment
Area of right triangle $= \dfrac{1}{2} \times 10 \times 10 = 50$ cm²
$$\text{Area of minor segment} = 78.5 - 50 = \mathbf{28.5 \text{ cm}^2}$$(ii) Major sector
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 the arc
(ii) Area of the sector
(iii) Area of the segment
Since the angle at centre is 60° and OA = OB = 21 cm (radii), △OAB is equilateral.
$$\text{Area of triangle} = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4} \text{ cm}^2$$$$\text{Area of segment} = 231 - \frac{441\sqrt{3}}{4} = \left(231 - \frac{441\sqrt{3}}{4}\right) \text{ cm}^2$$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 π = 3.14 and √3 = 1.73) #
Answer
Since OA = OB = 15 cm and ∠AOB = 60°, △OAB is equilateral.
Area of sector (60°):
$$= \frac{60}{360} \times 3.14 \times 225 = \frac{1}{6} \times 706.5 = 117.75 \text{ cm}^2$$Area of equilateral triangle:
$$= \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73 \times 225}{4} = \frac{389.25}{4} = 97.3125 \text{ cm}^2$$Minor segment $= 117.75 - 97.3125 = \mathbf{20.4375 \text{ cm}^2}$
Major segment $= 3.14 \times 225 - 20.4375 = 706.5 - 20.4375 = \mathbf{686.0625 \text{ cm}^2}$
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 π = 3.14 and √3 = 1.73) #
Answer
Area of sector (120°):
$$= \frac{120}{360} \times 3.14 \times 144 = \frac{1}{3} \times 3.14 \times 144 = 3.14 \times 48 = 150.72 \text{ cm}^2$$Area of triangle (OA = OB = 12 cm, ∠AOB = 120°):
$$= \frac{1}{2} \times 12 \times 12 \times \sin 120° = 72 \times \frac{\sqrt{3}}{2} = 36\sqrt{3} = 36 \times 1.73 = 62.28 \text{ cm}^2$$Area of segment $= 150.72 - 62.28 = \mathbf{88.44 \text{ cm}^2}$
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 π = 3.14) #
(i) Area with 5 m rope
The horse can graze a quadrant of radius 5 m (corner angle = 90°):
$$\text{Area} = \frac{90}{360} \times 3.14 \times 25 = \frac{1}{4} \times 78.5 = \mathbf{19.625 \text{ m}^2}$$(ii) Increase with 10 m rope
Area with 10 m rope $= \dfrac{1}{4} \times 3.14 \times 100 = 78.5$ m²
Increase $= 78.5 - 19.625 = \mathbf{58.875 \text{ m}^2}$
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 as shown in the figure. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch. #
(i) Total length of silver wire
Circumference $= \pi d = \dfrac{22}{7} \times 35 = 110$ mm
Length of 5 diameters $= 5 \times 35 = 175$ mm
Total length $= 110 + 175 = \mathbf{285 \text{ mm}}$
(ii) Area of each sector
10 equal sectors → each sector angle $= 36°$, radius $= \dfrac{35}{2}$ mm
$$\text{Area} = \frac{36}{360} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4} = \frac{22 \times 1225}{280} = \frac{26950}{280} = \frac{385}{4} \text{ mm}^2$$∴ Area of each sector $= \dfrac{385}{4}$ mm²
10. An umbrella has 8 ribs which are equally spaced (see the figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. #
Answer
8 ribs → angle between consecutive ribs $= \dfrac{360°}{8} = 45°$
$$\text{Area} = \frac{45}{360} \times \frac{22}{7} \times 45^2 = \frac{1}{8} \times \frac{22}{7} \times 2025$$$$= \frac{22 \times 2025}{56} = \frac{44550}{56} = \frac{22275}{28} \text{ cm}^2$$∴ Area $= \dfrac{22275}{28}$ cm²
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. #
Answer
Each wiper sweeps a sector of radius 25 cm and angle 115°.
$$\text{Area swept by one wiper} = \frac{115}{360} \times \frac{22}{7} \times 25^2 = \frac{115 \times 22 \times 625}{360 \times 7} = \frac{1581250}{2520} = \frac{158125}{252} \text{ cm}^2$$$$\text{Total area (two wipers)} = 2 \times \frac{158125}{252} = \frac{158125}{126} \text{ cm}^2$$∴ Total area cleaned $= \dfrac{158125}{126}$ cm²
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 π = 3.14) #
Answer
13. A round table cover has six equal designs as shown in the 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 √3 = 1.7) #
Answer
6 equal designs with central angle $= 60°$ each.
Area of each equilateral triangle (OA = OB = 28 cm):
$$= \frac{\sqrt{3}}{4} \times 784 = \frac{1.7 \times 784}{4} = 333.2 \text{ cm}^2$$Area of each sector:
$$= \frac{60}{360} \times \frac{22}{7} \times 784 = \frac{1232}{3} = 410.67 \text{ cm}^2$$Area of each segment $= 410.67 - 333.2 = 77.47$ cm²
Total design area $= 6 \times 77.47 = 464.8$ cm²
Cost $= 464.8 \times 0.35 = \mathbf{₹162.68}$
14. Tick the correct answer in the following: Area of a sector of angle $p$ (in degrees) of a circle with radius $R$ is: (A) $\dfrac{p}{180} \times 2\pi R$ (B) $\dfrac{p}{180} \times \pi R^2$ (C) $\dfrac{p}{360} \times 2\pi R$ (D) $\dfrac{p}{720} \times 2\pi R^2$ #
Answer
Area of sector $= \dfrac{p}{360} \times \pi R^2 = \dfrac{p}{360} \times \pi R^2$
Checking option D: $\dfrac{p}{720} \times 2\pi R^2 = \dfrac{2p\pi R^2}{720} = \dfrac{p\pi R^2}{360}$ ✓
Answer: (D) $\dfrac{p}{720} \times 2\pi R^2$