CBSE Class 10 Areas related to Circles Sure Shot Questions Set B

Question. If the sum of the areas of two circles with radii \( R_1 \) and \( R_2 \) is equal to the area of a circle of radius \( R \), then
(a) \( R_1 + R_2 = R \)
(b) \( R_1^2 + R_2^2 = R^2 \)
(c) \( R_1 + R_2 < R \)
(d) \( R_1^2 + R_2^2 < R^2 \)
Answer: (b)
Explanation: According to given condition,
Area of circle = Area of first circle + Area of second circle
\( \pi R^2 = \pi R_1^2 + \pi R_2^2 \)
\( R^2 = R_1^2 + R_2^2 \)

Question. If the circumference of a circle and the perimeter of a square are equal, then
(a) Area of the circle = Area of the square
(b) Area of the circle > Area of the square
(c) Area of the circle < Area of the square
(d) Nothing definite can be said about the relation between the areas of the circle and square.
Answer: (b)
Explanation: According to given condition Circumference of a circle = Perimeter of square.
\( 2\pi r = 4a \)
[Where \( r \) and \( a \) are radius of circle and side of square respectively]
\( \frac{22}{7} r = 2a \)
\( 11r = 7a \)
\( r = \frac{7a}{11} \) ...(1)
Area of circle, \( A_1 = \pi r^2 \)
\( = \pi \left( \frac{7a}{11} \right)^2 \)
\( = \frac{14a^2}{11} \) ......(2)
Area of square, \( A_2 = a^2 \) ......(3)
From equations (2) and (3)
\( A_1 > A_2 \)
Hence Area of the circle > Area of the square.

Question. Area of the largest triangle that can be inscribed in a semi-circle of radius \( r \) units, in square units is:
(a) \( r^2 \)
(b) \( 1/2r^2 \)
(c) \( 2r^2 \)
(d) \( \sqrt{2}r^2 \)
Answer: (a)
Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.
We know that,
Area of a triangle \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\(\therefore\) Area of \( \Delta ABC = \frac{1}{2} \times AC \times BD \)
\( \Rightarrow \text{Area of } \Delta ABC = \frac{1}{2} \times 2r \times r \)
\( \Rightarrow \text{Area of } \Delta ABC = r^2 \)

Question. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is:
(a) 22:7
(b) 14:11
(c) 7:22
(d) 11:14
Answer: (b)
Explanation: Perimeter of circle = Perimeter of square
\( 2\pi r = 4a \)
\( \Rightarrow a = \frac{\pi r}{2} \)
\( \frac{\text{Area of circle}}{\text{Area of square}} = \frac{\pi r^2}{(\frac{\pi r}{2})^2} \)
\( \Rightarrow \frac{\text{Area of circle}}{\text{Area of square}} = \frac{4}{\frac{22}{7}} \)
\( \Rightarrow \frac{\text{Area of circle}}{\text{Area of square}} = \frac{14}{11} \)

Question. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer: (a)
Explanation: Area of first circular park, whose diameter is 16m
\( = \pi r^2 = \pi (16/2)^2 = 64\pi \text{ m}^2 \)
Area of second circular park, whose diameter is 12m
\( = \pi r^2 = \pi (12/2)^2 = 36\pi \text{ m}^2 \)
According to question,
Area of new circular park = \( (64\pi + 36\pi) \text{ m}^2 \)
\( \pi R^2 = 100\pi \text{ m}^2 \)
\( R = 10\text{m} \)

Question. The area of the circle that can be inscribed in a square of side 6 cm is
(a) \( 36\pi \text{ cm}^2 \)
(b) \( 18\pi \text{ cm}^2 \)
(c) \( 12\pi \text{ cm}^2 \)
(d) \( 9\pi \text{ cm}^2 \)
Answer: (d)
Explanation: Given, Side of square = 6 cm
Diameter of a circle = side of square = 6cm
Therefore, Radius of circle = 3cm
Area of circle \( = \pi r^2 = \pi (3)^2 = 9\pi \text{ cm}^2 \)

Question. The area of the square that can be inscribed in a circle of radius 8 cm is
(a) \( 256\text{ cm}^2 \)
(b) \( 128\text{ cm}^2 \)
(c) \( 642\text{ cm}^2 \)
(d) \( 64\text{ cm}^2 \)
Answer: (b)
Explanation: Radius of circle = 8 cm
Diameter of circle = 16 cm = diagonal of the square
Therefore side of square \( = \text{diagonal}/\sqrt{2} = 16/\sqrt{2} \)
Therefore, area of square is \( = (\text{side})^2 = (16/\sqrt{2})^2 = 256/2 = 128\text{ cm}^2 \)

Question. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer: (c)
Explanation: According to question, Circumference of circle = Circumference of first circle + Circumference of second circle
\( \pi D = \pi d_1 + \pi d_2 \)
\( D = 36 + 20 \)
\( D = 56\text{cm} \)
So, Radius \( = 56/2 = 28\text{cm} \)

Question. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm respectively, is
(a) 31 cm
(b) 25 cm
(c) 62 cm
(d) 50 cm
Answer: (d)
Explanation: According to question
\( \pi R^2 = \pi r_1^2 + \pi r_2^2 \)
\( \pi R^2 = [\pi (24)^2 + \pi (7)^2] \text{ cm}^2 \)
\( R^2 = (576 + 49) \text{ cm}^2 \)
\( R^2 = 625 \text{ cm}^2 \)
\( R = 25 \text{ cm} \)
Therefore diameter \( = 2 \times 25 = 50\text{cm} \)

Question. If the length of an arc of a circle of radius \( r \) is equal to that of an arc of a circle of radius \( 2r \), then
(a) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.
(b) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.
(c) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.
(d) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.
Answer: (a)
Explanation: According to Question,
\( \frac{\theta_1}{360} \times 2\pi r_1 = \frac{\theta_2}{360} \times 2\pi r_2 \)
\( \theta_1 r_1 = \theta_2 r_2 \)
\( \theta_1 r = \theta_2 (2r) \)
\( \theta_1 = 2\theta_2 \)

Question. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?
(a) 300
(b) 400
(c) 450
(d) 500
Answer: (d)
Explanation:
Circumference of the wheel \( = 2\pi r \)
\( = 2 \times \frac{22}{7} \times 35 \)
\( = 220\text{cm} \)
Speed of the wheel \( = 66\text{ km/hr} \)
\( = \frac{66 \times 1000}{60} = 1100 \times 100 \text{ cm/min} \)
\( = 110000 \text{ cm/min} \)
Number of revolutions per minute \( = \frac{110000}{220} = 500 \)

Question. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:
(a) \( 154\text{ m}^2 \)
(b) \( 156\text{ m}^2 \)
(c) \( 158\text{ m}^2 \)
(d) \( 160\text{ m}^2 \)
Answer: (a)
Explanation: Figure according to question is:
Area of the field in which cow can graze = Area of a sector AFEG
\( = (\theta/360) \times \pi r^2 \)
\( = (90/360) \times \pi (14)^2 \)
\( = (1/4) \times (22/7) \times 196 \)
\( = 154\text{ m}^2 \)

Question. Area of a sector of central angle 120° of a circle is \( 3\pi \text{ cm}^2 \). Then the length of the corresponding arc of this sector is:
(a) 5.8cm
(b) 6.1cm
(c) 6.3cm
(d) 6.8cm
Answer: (c)
Explanation: Given that Area of a sector of central angle 120° of a circle is \( 3\pi \text{ cm}^2 \)
Area of sector \( = \frac{\theta}{360} \pi r^2 \)
\( \Rightarrow 3\pi = \frac{120}{360} \pi r^2 \)
\( \Rightarrow 3 = \frac{1}{3} r^2 \)
\( \Rightarrow r^2 = 9 \)
\( \Rightarrow r = 3 \text{ cm} \)
Therefore, required length of arc \( = \frac{\theta}{360} 2\pi r \)
\( = \frac{120}{360} \cdot 2 \cdot \frac{22}{7} \cdot 3 \)
\( = 2 \cdot \frac{22}{7} \)
\( = 6.3\text{cm} \)

Very Short Answer

Question. If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the larger circle (in cm).
Answer: Sol. \( R^2 = r_1^2 + r_2^2 \)
\( \pi R^2 = \pi(r_1^2 + r_2^2) \)
\( r_1 = \frac{10}{2} = 5 \) cm, \( r_2 = \frac{24}{2} = 12 \) cm
\( R^2 = 5^2 + 12^2 = 25 + 144 \)
\( R^2 = 169 \Rightarrow R = 13 \) cm
\( \therefore \) Diameter \( = 2(13) = 26 \) cm

Question. The circumference of a circle is 22 cm. Calculate the area of its quadrant (in cm\(^2\)).
Answer: Sol. Circumference of a circle \( = 22 \) cm \( \Rightarrow 2\pi r = 22 \) cm
\( 2 \times \frac{22}{7} \times r = 22 \) cm
\( r = \frac{22 \times 7}{22 \times 2} = \frac{7}{2} \) cm
\( \therefore \) Area of quadrant \( = \frac{1}{4} \pi r^2 \)
\( = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \) cm\(^2\)

Question. If the difference between the circumference and the radius of a circle is 37 cm, then using \( \pi = \frac{22}{7} \), calculate the circumference (in cm) of the circle.
Answer: Sol. \( 2\pi r - r = 37 \Rightarrow r(2\pi - 1) = 37 \)
\( \Rightarrow r\left(\frac{44}{7} - 1\right) = 37 \Rightarrow r\left(\frac{37}{7}\right) = 37 \)
\( \Rightarrow r = 37 \times \frac{7}{37} = 7 \) cm
Circumference of the circle \( = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \) cm

Question. If \( \pi \) is taken as \( \frac{22}{7} \), calculate the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution.
Answer: Sol. Radius \( (r) = \frac{35}{2} \)
Required distance \( = \text{Perimeter} = 2\pi r \)
\( = 2 \times \frac{22}{7} \times \frac{35}{2} \) cm \( = 110 \) cm or \( 1.1 \) m

Short Answer

Question. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer: Sol. Here \( \theta = \frac{360^\circ}{60 \text{ m}} \times 5 \text{ m} = 30^\circ \) ...[\( \because \) 1 hour \( = 60 \) minutes]
\( r(\text{radius}) = 14 \) cm
\( \therefore \) Required area \( = \frac{\theta}{360} \pi r^2 \)
\( = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14 \)
\( = \frac{154}{3} \text{ cm}^2 \) or \( 51.3 \text{ cm}^2 \)

Question. Find the area of a quadrant of a circle, where the circumference of circle is 44 cm. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Circumference of a circle \( = 44 \) cm
\( \Rightarrow 2\pi r = 44 \) cm
\( \Rightarrow 2 \times \frac{22}{7} \times r = 44 \) cm
\( \Rightarrow r = \frac{44 \times 7}{2 \times 22} = 7 \) cm
Area of a quadrant \( = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 \)
\( \therefore \) Area of quadrant \( = \frac{77}{2} = 38.5 \text{ cm}^2 \)

Question. Area of a sector of a circle of radius 14 cm is 154 cm\(^2\). Find the length of the corresponding arc of the sector. [Use \( \pi = \frac{22}{7} \)]
Answer: Sol. Area of sector \( = 154 \text{ cm}^2 \)
\( \Rightarrow \frac{1}{2} lr = 154 \Rightarrow \frac{1}{2} (l)(14) = 154 \)
\( \Rightarrow 7l = 154 \Rightarrow l = 22 \) cm
\( \therefore \) Length of the corresponding arc, \( l = 22 \) cm

Question. Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15 cm and 18 cm.
Answer: Sol. Let the radius of bigger, first and second circles are \( r, r_1 \) and \( r_2 \) respectively.
According to the question, Circumference of bigger circle = Circumference of first circle + Circumference of second circle
\( 2\pi r = 2\pi r_1 + 2\pi r_2 \Rightarrow r = r_1 + r_2 \)
\( r = 15 + 18 = 33 \text{ cm} \) ...[Given \( r_1 = 15 \text{ cm}, r_2 = 18 \text{ cm} \)]
Hence, required radius of circle is 33 cm.

Question. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make, so as to keep a speed of 66 km/h?
Answer: Sol. Given. Radius of wheel, \( r = 35 \) cm
Speed of wheel = 66 km/h \( = \frac{66 \times 1000}{60} \times 100 \text{ cm/min} \)
Thus, distance covered by wheel in one min. = 110000 cm
\( \therefore \) Distance covered by wheel in one revolution \( = 2\pi r = 2 \times \frac{22}{7} \times 35 = 220 \text{ cm} \)
\( \therefore \) Number of revolutions to cover 1,10,000 cm \( = \frac{110000}{220} = 500 \text{ revolutions} \)
Hence, the wheel completes 500 revolutions per minute.

Long Answer Type Questions

Question. The area of a circular playground is 22,176 m\(^2\). Find the cost of fencing this ground at the rate of ₹50 per m.
Answer: Sol. Area of a circular playground = 22,176 m\(^2\) \( \Rightarrow \pi r^2 = 22,176 \Rightarrow \frac{22}{7} r^2 = 22,176 \Rightarrow r^2 = \frac{22176 \times 7}{22} \Rightarrow r^2 = 7056 \Rightarrow r = \sqrt{7056} = 84 \text{ m} \).
Now, circumference of a circle \( = 2\pi r = 2 \times \frac{22}{7} \times 84 = 44 \times 12 = 528 \text{ m} \).
\( \therefore \text{Cost of fencing the playground @ ₹50 per m} = 528 \times ₹50 = ₹26,400 \)

Question. The diameters of front and rear wheels of a tractor are 80 cm and 200 cm, respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1,400 revolutions.
Answer: Sol. Radius of front wheel, \( r_1 = \frac{80}{2} = 40 \text{ cm} \), and radius of rear wheel, \( r_2 = \frac{200}{2} = 100 \text{ cm} \).
Now, circumference of the front wheel \( = 2\pi r_1 = 2 \times \frac{22}{7} \times 40 = \frac{1760}{7} \).
Total distance covered by front wheel \( = 1,400 \times \frac{1760}{7} = 3,52,000 \text{ cm} \).
As, distance travelled by rear wheel = Distance travelled by front wheel, and circumference of the rear wheel \( = 2\pi r_2 = 2 \times \frac{22}{7} \times 100 = \frac{4400}{7} \).
\( \therefore \text{No. of revolutions made by rear wheel} = \frac{\text{Distance covered}}{\text{Circumference}} = \frac{352000 \times 7}{4400} = 560 \text{ revolutions} \)

Question. The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 am.
Answer: Sol. Time swept = 35 min. Now, angle revolved in 60 minutes = 360°. \( \therefore \) Angle revolved in 35 minutes, \( \theta = \frac{360^\circ}{60^\circ} \times 35 = 210^\circ \).
\( \text{Required area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{210^\circ}{360^\circ} \times \frac{22}{7} \times (5)^2 = \frac{25 \times 22}{12} = 45.83 \text{ cm}^2 \). Hence, the required area is 45.83 cm\(^2\).

Question. Area of a sector of central angle 200° of a circle is 770 cm\(^2\). Find the length of the corresponding arc of this sector.
Answer: Sol. Given, central angle \( \theta = 200^\circ \) and area of sector = 770 cm\(^2\).
\( \Rightarrow \theta \times \frac{\pi r^2}{360^\circ} = 770 \Rightarrow r^2 = \frac{360 \times 770}{\pi \times \theta} = \frac{360 \times 770 \times 7}{22 \times 200} \Rightarrow r^2 = 441 \Rightarrow r = 21 \text{ cm} \).
Now, length of corresponding arc, \( l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{200 \times 2 \times 22}{360^\circ \times 7} \times 21 = \frac{220}{3} = 73.3 \text{ cm} \).

Question. The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?
Answer: Sol. \( A_1 = \frac{\theta_1}{360^\circ} \pi r_1^2 = \frac{120^\circ}{360^\circ} \times \frac{22}{7} \times 7 \times 7 = \frac{154}{3} \text{ cm}^2 \).
\( A_2 = \frac{\theta_2}{360^\circ} \pi r_2^2 = \frac{40^\circ}{360^\circ} \times \frac{22}{7} \times 21 \times 21 = 154 \text{ cm}^2 \).
\( l_1 = \frac{\theta_1}{360^\circ} 2\pi r_1 = \frac{120^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{44}{3} \text{ cm} \).
\( l_2 = \frac{\theta_2}{360^\circ} 2\pi r_2 = \frac{40^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{44}{3} \text{ cm} \).
Observation: arc lengths are equal but area of second sector is three times that of the first sector.

Question. Find the number of revolutions made by a circular wheel of area 1.54 m\(^2\) in rolling a distance of 176 m.
Answer: Sol. \( \pi r^2 = 1.54 \Rightarrow r^2 = \frac{1.54}{22} \times 7 \Rightarrow r^2 = 0.49 \Rightarrow r = 0.7 \text{ m} \).
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 0.7 = 4.4 \text{ m} \).
Number of revolutions \( = \frac{176}{4.4} = 40 \).

Facts that Matter

• Perimeter and Area of a Circle
• The total distance (perimeter) around a circle is called its circumference.
• The plane surface enclosed in a circle is called its area.
• If ‘r’ be the radius of a circle, then
  • (i) Circumference of the circle = \(2\pi r\)
  • (ii) Area of the circle = \(\pi r^2\)

NOTE:
I. The interior of a circle along with its boundary is called the circular region of the circle.
II. By the area of a circle, we mean the area of the circular region.

Question. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Answer: We have, \(r_1 = 19 \text{ cm}\)
\(r_2 = 9 \text{ cm}\)
\(\therefore\) Circumference of circle-I = \(2\pi r_1 = 2\pi (19) \text{ cm}\)
Circumference of circle-II = \(2\pi r_2 = 2\pi (9) \text{ cm}\)
Sum of the circumferences of circle-I and circle-II
= \(2\pi (19) + 2\pi (9) = 2\pi (19 + 9) \text{ cm} = 2\pi (28) \text{ cm}\)
Let \(R\) be the radius of the circle-III.
\(\therefore\) Circumference of circle-III = \(2\pi R\)
According to the condition,
\(2\pi R = 2\pi (28)\)
\(\Rightarrow R = \frac{2\pi(28)}{2\pi} = 28 \text{ cm}\)
Thus, the radius of the new circle = 28 cm.

Question. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Answer: We have,
Radius of circle-I, \(r_1 = 8 \text{ cm}\)
Radius of circle-II, \(r_2 = 6 \text{ cm}\)
\(\therefore\) Area of circle-I = \(\pi r_1^2 = \pi (8)^2 \text{ cm}^2\)
Area of circle-II = \(\pi r_2^2 = \pi (6)^2 \text{ cm}^2\)
Let the area of the circle-III be \(R\)
\(\therefore\) Area of circle-III = \(\pi R^2\)
Now, according to the condition,
\(\pi r_1^2 + \pi r_2^2 = \pi R^2\)
i.e. \(\pi (8)^2 + \pi (6)^2 = \pi R^2\)
\(\Rightarrow \pi (8^2 + 6^2) = \pi R^2\)
\(\Rightarrow 8^2 + 6^2 = R^2\)
\(\Rightarrow 64 + 36 = R^2\)
\(\Rightarrow 100 = R^2\)
\(\Rightarrow 10^2 = R^2 \Rightarrow R = 10\)
Thus, the radius of the new circle = 10 cm.

Question. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
Answer: Diameter of a wheel = 80 cm
\(\therefore\) Radius of the wheel = \(\frac{80}{2} = 40 \text{ cm}\)
\(\therefore\) Circumference of the wheel
= \(2\pi \times 40 = 2 \times \frac{22}{7} \times 40 \text{ cm}\)
\(\Rightarrow\) Distance covered by a wheel in one revolution = \(\frac{2 \times 22 \times 40}{7} \text{ cm}\)
Distance travelled by the car in 1hr
= 66 km = 66 \(\times\) 1000 \(\times\) 100 cm
\(\therefore\) Distance travelled in 10 minutes
= \(\frac{66 \times 1000 \times 100}{60} \times 10 \text{ cm} = 11 \times 100000 \text{ cm}\)
Now,
Number of revolutions
= \(\frac{\text{Distance travelled in 10 minutes}}{\text{Distance travelled in one revolution}}\)
= \(\frac{1100000}{\left[ \frac{2 \times 22 \times 40}{7} \right]} = \frac{1100000 \times 7}{2 \times 22 \times 40} = 4375\)
Thus, the required number of revolutions = 4375.

Area of Sector and Segment of a Circle

The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle.

The portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle.

NOTE:
I. \(\angle AOB\) is called the ‘angle of sector’.
II. OAPB is the ‘minor sector’ and OAQB is the ‘major sector’.
III. APB is the ‘minor segment’ and AQB is the ‘major segment’.
IV. When we write ‘sector’ and ‘segment’ we will mean the ‘minor-sector’ and the ‘minor segment’ respectively.

Let us remember that
(i) Area of the sector of ‘angle \(\theta\) = \(\frac{\theta}{360^\circ} \times \pi r^2\)
(ii) Length of the arc of a sector of angle \(\theta\) = \(\frac{\theta}{360^\circ} \times 2\pi r\)
(iii) Area of a segment = [Area of the corresponding sector] − [Area of the corresponding triangle]

Question. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer: [Length of minute hand] = [radius of the circle]
\(\Rightarrow r = 14 \text{ cm}\)
\(\because\) Angle swept by the minute hand in 60 minutes = \(360^\circ\)
\(\therefore\) Angle swept by the minute hand in 5 minutes = \(\frac{360^\circ}{60} \times 5 = 30^\circ\)
Now, area of the sector with \(r = 14 \text{ cm}\) and \(\theta = 30^\circ\)
\(\frac{\theta}{360} \times \pi r^2 = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14 \text{ cm}^2 = \frac{11 \times 14}{3} \text{ cm}^2 = \frac{154}{3} \text{ cm}^2\)
Thus, the required area swept by the minute hand by 5 minutes = \(\frac{154}{3} \text{ cm}^2\).

Question. 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 \(\pi = 3.14\))
Answer: Length of the radius (r) = 10 cm
Sector angle \(\theta = 90^\circ\)
Area of the sector with \(\theta = 90^\circ\) and \(r = 10 \text{ cm}\)
= \(\frac{90}{360} \times 3.14 \times 10 \times 10 \text{ cm}^2 = \frac{1}{4} \times 314 \text{ cm}^2 = \frac{157}{2} \text{ cm}^2 = 78.5 \text{ cm}^2\)
Now,
(i) Area of the minor segment
= [Area of minor sector] − [Area of rt. \(\Delta AOB\)]
= [78.5 \(\text{cm}^2\)] − \([\frac{1}{2} \times 10 \times 10] \text{ cm}^2 = 78.5 \text{ cm}^2 - 50 \text{ cm}^2 = 28.5 \text{ cm}^2\).
(ii) Area of major segment
= [Area of the circle] − [Area of the minor segment]
= \(\pi r^2 - 78.5 \text{ cm}^2\)
= \(\left[ \frac{314}{100} \times 10 \times 10 - 78.5 \right] \text{ cm}^2 = (314 - 78.5) \text{ cm}^2 = 235.5 \text{ cm}^2\).

Question. 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.
Answer: Here, radius (r) = 25 cm
Sector angle (\(\theta\)) = \(115^\circ\)
\(\therefore\) Area cleaned by each sweep of the blades
= \(\left[ \frac{\theta}{360} \times \pi r^2 \right] \times 2\) [\(\because\) Each sweep will have to and fro movement]
= \(\frac{115}{360} \times \frac{22}{7} \times 25 \times 25 \times 2 \text{ cm}^2\)
= \(\frac{23 \times 11 \times 25 \times 25}{18 \times 7} \text{ cm}^2 = \frac{158125}{126} \text{ cm}^2\).

Question. 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 \(\pi = 3.14\))
Answer: Here, Radius (r) = 16.5 km
Sector angle (\(\theta\)) = \(80^\circ\)
\(\therefore\) Area of the sea surface over which the ships are warned
= \(\frac{\theta}{360} \times \pi r^2\)
= \(\frac{80}{360} \times 3.14 \times \frac{165}{10} \times \frac{165}{10} \text{ km}^2\)
= \(\frac{157 \times 11 \times 11}{100} \text{ km}^2 = \frac{18997}{100} \text{ km}^2 = 189.97 \text{ km}^2\).

Question. Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is
(a) \(\frac{p}{180} \times 2\pi R\)
(b) \(\frac{p}{180} \times \pi R^2\)
(c) \(\frac{p}{360} \times 2\pi R\)
(d) \(\frac{p}{720} \times 2\pi R^2\)
Answer: (d) Here, radius (r) = R, Angle of sector (\(\theta\)) = \(p^\circ\)
Area of sector = \(\frac{\theta}{360} \times \pi r^2 = \frac{p}{360} \times \pi R^2 = \frac{p}{720} \times 2\pi R^2\).