NCERT Solutions For Class 10 Maths: Area Related to Circles

EXERCISE 11.1

Question 1: Find the area of a sector of a circle with radius 6 cm if angle of the sector is $60^\circ$.

Solution-
1.Given radius $r = 6$ cm and angle of the sector $\theta = 60^\circ$.

2.The Area of the sector of angle $\theta$ is given by the formula $\frac{\theta}{360} \times \pi r^2$.

3.Area $= \frac{60}{360} \times \frac{22}{7} \times (6)^2$

4.Area $= \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{22 \times 6}{7} = \frac{132}{7}$ cm$^2$.

Question 2: Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution-
1.The circumference of the circle is $2\pi r = 22$ cm.

2.Find the radius $r$:
$2 \times \frac{22}{7} \times r = 22$
$r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2}$ cm.

3.A quadrant of a circle is a sector with an angle $\theta = 90^\circ$.

4.Area of the quadrant $= \frac{90}{360} \times \pi r^2 = \frac{1}{4} \pi r^2$.

5.Area $= \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4}$

6.Area $= \frac{11 \times 7}{8} = \frac{77}{8}$ cm$^2$.

Question 3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution-
1.The length of the minute hand acts as the radius, $r = 14$ cm.

2.In 60 minutes, the minute hand sweeps $360^\circ$. In 1 minute, it sweeps $\frac{360}{60} = 6^\circ$.

3.The angle swept in 5 minutes is $\theta = 5 \times 6^\circ = 30^\circ$.

4.The area swept is the Area of the sector formed: $\frac{\theta}{360} \times \pi r^2$.

5.Area $= \frac{30}{360} \times \frac{22}{7} \times (14)^2 = \frac{1}{12} \times \frac{22}{7} \times 196$

6.Area $= \frac{1}{12} \times 22 \times 28 = \frac{154}{3}$ cm$^2$.

Question 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 $\pi = 3.14$)

Solution-
1.Given radius $r = 10$ cm and angle of the sector $\theta = 90^\circ$.

(i) Area of the minor segment:
1.Area of minor sector $= \frac{90}{360} \times \pi r^2$.
Area of minor sector $= \frac{1}{4} \times 3.14 \times (10)^2 = 0.25 \times 314 = 78.5$ cm$^2$.

2.Area of $\triangle OAB$ (corresponding triangle): Since $\theta = 90^\circ$, Area $= \frac{1}{2} \times r \times r$.
Area of $\triangle OAB = \frac{1}{2} \times 10 \times 10 = 50$ cm$^2$.

3.Area of minor segment = Area of sector OAPB – Area of $\triangle OAB$.
Area $= 78.5 - 50 = 28.5$ cm$^2$.

(ii) Area of the major sector:
1.Area of circle $= \pi r^2 = 3.14 \times 100 = 314$ cm$^2$.

2.Area of major sector $= \pi r^2$ – Area of the minor sector OAPB.
Area $= 314 - 78.5 = 235.5$ cm$^2$.

Question 5: In a circle of radius 21 cm, an arc subtends an angle of $60^\circ$ 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

Solution-
1.Given $r = 21$ cm and angle $\theta = 60^\circ$.

(i) The length of the arc:
1.Length of an arc of a sector $= \frac{\theta}{360} \times 2\pi r$.
Length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 44 \times 3 = 22$ cm.

(ii) Area of the sector formed by the arc:
1.Area $= \frac{\theta}{360} \times \pi r^2$.
Area $= \frac{60}{360} \times \frac{22}{7} \times 21^2 = \frac{1}{6} \times \frac{22}{7} \times 441 = 231$ cm$^2$.

(iii) Area of the segment formed by the corresponding chord:
1.Area of segment = Area of sector – Area of $\triangle OAB$.
Since $OA=OB$ and $\theta=60^\circ$, $\triangle OAB$ is an equilateral triangle.
Area of $\triangle OAB = \frac{\sqrt{3}}{4} r^2 = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4}$ cm$^2$.

2.Area of segment $= \left(231 - \frac{441\sqrt{3}}{4}\right)$ cm$^2$.

Question 6: A chord of a circle of radius 15 cm subtends an angle of $60^\circ$ at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution-
1.Given $r = 15$ cm and $\theta = 60^\circ$.

(i) Area of the minor segment:
1.Area of minor sector $= \frac{60}{360} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = 117.75$ cm$^2$.

2.Area of the corresponding triangle $\triangle OAB$ (equilateral, since $\theta=60^\circ$):
Area $= \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 15^2 = \frac{1.73 \times 225}{4} = 97.3125$ cm$^2$.

3.Area of minor segment $= 117.75 - 97.3125 = 20.4375$ cm$^2$.

(ii) Area of the major segment:
1.Area of circle $= \pi r^2 = 3.14 \times 15^2 = 706.5$ cm$^2$.

2.Area of major segment $= \pi r^2$ – Area of the minor segment.
Area $= 706.5 - 20.4375 = 686.0625$ cm$^2$.

Question 7: A chord of a circle of radius 12 cm subtends an angle of $120^\circ$ at the centre. Find the area of the corresponding segment of the circle.
(Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution-
1.Given $r = 12$ cm and $\theta = 120^\circ$. We find the area of the minor segment.

2.Area of the sector OAPB $= \frac{120}{360} \times \pi r^2 = \frac{1}{3} \times 3.14 \times 12^2 = 150.72$ cm$^2$.

3.Area of the corresponding triangle $\triangle OAB$:
Area $= \frac{1}{2} r^2 \sin \theta$.
Area $= \frac{1}{2} \times 12^2 \times \sin 120^\circ = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2}$

4.Area of $\triangle OAB = 36\sqrt{3} = 36 \times 1.73 = 62.28$ cm$^2$.

5.Area of segment $= 150.72 - 62.28 = 88.44$ cm$^2$.

Question 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 (see Fig. 11.8). 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 $\pi = 3.14$)

Solution-
1.Since the field is square shaped, the angle at the corner where the horse is tied is $90^\circ$. The grazing area is a sector with $\theta = 90^\circ$.

(i) Area grazed with $r_1 = 5$ m rope:
1.Area $= \frac{90}{360} \times \pi r_1^2 = \frac{1}{4} \times 3.14 \times 5^2$.

2.Area $= \frac{3.14 \times 25}{4} = 19.625$ m$^2$.

(ii) The increase in the grazing area if the rope were $r_2 = 10$ m:
1.Area grazed with $r_2 = 10$ m:
Area$_2 = \frac{1}{4} \times 3.14 \times 10^2 = \frac{314}{4} = 78.5$ m$^2$.

2.Increase in area $= \text{Area}_2 - \text{Area}_1 = 78.5 - 19.625 = 58.875$ m$^2$.

Question 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 Fig. 11.9. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.

Solution-
1.Given Diameter $D = 35$ mm, so radius $r = 35/2 = 17.5$ mm.

(i) The total length of the silver wire required:
1.Total Length = Circumference of circle + Length of 5 diameters.

2.Circumference $= \pi D = \frac{22}{7} \times 35 = 110$ mm.

3.Length of 5 diameters $= 5 \times 35 = 175$ mm.

4.Total length $= 110 + 175 = 285$ mm.

(ii) The area of each sector of the brooch:
1.The circle is divided into 10 equal sectors. The angle of each sector is $\theta = \frac{360^\circ}{10} = 36^\circ$.

2.Area of one sector $= \frac{1}{10} \times \pi r^2$.
Area $= \frac{1}{10} \times \frac{22}{7} \times \left(\frac{35}{2}\right)^2 = \frac{1}{10} \times \frac{22}{7} \times \frac{1225}{4}$

3.Area $= \frac{385}{4} = 96.25$ mm$^2$.

Question 10: An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution-
1.The umbrella forms 8 equally spaced sectors of a flat circle with radius $r = 45$ cm.

2.The angle between two consecutive ribs is $\theta = \frac{360^\circ}{8} = 45^\circ$.

3.The area between the two consecutive ribs is the Area of one sector: $\frac{\theta}{360} \times \pi r^2$.

4.Area $= \frac{45}{360} \times \frac{22}{7} \times 45^2 = \frac{1}{8} \times \frac{22}{7} \times 2025$

5.Area $= \frac{11 \times 2025}{4 \times 7} = \frac{22275}{28}$ cm$^2$.

Question 11: A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of $115^\circ$. Find the total area cleaned at each sweep of the blades.

Solution-
1.The length of the blade acts as the radius $r = 25$ cm. The angle of sweep is $\theta = 115^\circ$.

2.The area swept by one wiper is the Area of the sector: $\frac{\theta}{360} \times \pi r^2$.

3.Since there are two wipers that do not overlap, the total area cleaned is $2 \times \text{Area of one sector}$.
Total Area $= 2 \times \frac{115}{360} \times \frac{22}{7} \times 25^2$

4.Total Area $= 2 \times \frac{23}{72} \times \frac{22}{7} \times 625 = \frac{158125}{126}$ cm$^2$.

Question 12: To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle $80^\circ$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $\pi = 3.14$)

Solution-
1.The distance the light reaches acts as the radius, $r = 16.5$ km. The sector angle is $\theta = 80^\circ$.

2.The area of the sea warned is the Area of the sector: $\frac{\theta}{360} \times \pi r^2$.

3.Area $= \frac{80}{360} \times 3.14 \times (16.5)^2 = \frac{2}{9} \times 3.14 \times 272.25$

4.Area $\approx 189.97$ km$^2$.

Question 13: A round table cover has six equal designs as shown in Fig. 11.11. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of $₹ 0.35$ per cm$^2$. (Use $\sqrt{3} = 1.7$)

Solution-
1.Given radius $r = 28$ cm. The six equal designs are minor segments. Since there are 6 equal segments, the angle subtended by each chord at the centre is $\theta = \frac{360^\circ}{6} = 60^\circ$.

2.Area of one design (segment) = Area of sector – Area of corresponding triangle.

3.Area of sector:
Area $= \frac{60}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times 22 \times 4 \times 28 = \frac{1232}{3} \approx 410.67$ cm$^2$.

4.Area of triangle $\triangle OAB$: Since $r=28$ and $\theta=60^\circ$, it is an equilateral triangle.
Area $= \frac{\sqrt{3}}{4} r^2 = \frac{1.7}{4} \times 28^2 = 1.7 \times 7 \times 28 = 333.2$ cm$^2$.

5.Area of one design (segment) $= 410.67 - 333.2 = 77.47$ cm$^2$ (approx.).

6.Total area of 6 designs $= 6 \times 77.47 = 464.82$ cm$^2$.

7.Cost of making the designs $= \text{Total Area} \times \text{Rate} = 464.82 \times 0.35 = 162.68$ (in ₹).

Question 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) $\frac{p}{180} \times \pi R^2$ (B) $\frac{p}{180} \times 2\pi R$ (C) $\frac{p}{360} \times \pi R^2$ (D) $\frac{p}{720} \times 2\pi R^2$

Solution-
1.The Area of a sector with angle $\theta$ and radius $r$ is defined as $\frac{\theta}{360} \times \pi r^2$.

2.Substituting $p$ for $\theta$ and $R$ for $r$, the formula is $\frac{p}{360} \times \pi R^2$.

3.Option (D) is $\frac{p}{720} \times 2\pi R^2$. Since $\frac{2}{720} = \frac{1}{360}$, Option (D) simplifies to $\frac{p}{360} \times \pi R^2$.

4.The correct option is (D) $\frac{p}{720} \times 2\pi R^2$.