CBSE Class 10 Mathematics Areas Related to Circles Assignment Set 09

NCERT Exercise: Area of Sector and Segment of a Circle

Question 1. Find the area of a sector of a circle with radius 6 cm if the angle of the sector is \( 60^\circ \).
Answer:
Here \( r = 6 \) cm and \( \theta = 60^\circ \). Area of the sector \( = \frac{\theta}{360^\circ} \times \pi r^2 \).
\( = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 6 \times 6 = \frac{22}{7} \times 6 = \frac{132}{7} \) cm².
Hence, the area of the sector is \( \frac{132}{7} \) cm².
In simple words: A 60 degree sector is one-sixth of the whole circle.

Teacher's Note:
1. Write the formula first (sector area = angle/360 times pi r squared) and then substitute the values.
2. Always give the unit in cm², m² and so on.

Question 2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Answer:
Circumference \( = 2\pi r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{7}{2} \) cm.
A quadrant is a sector making an angle of \( 90^\circ \). Area of the quadrant \( = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \) cm².
Hence, the area of the quadrant is \( \frac{77}{8} \) cm².
In simple words: A quadrant is one-fourth of a circle.

Teacher's Note:
1. First find the radius from the circumference.
2. A quadrant means a sector of angle 90 degrees.

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.
Answer:
The minute hand is the radius of the circle: \( r = 14 \) cm. The minute hand turns \( 360^\circ \) in 60 minutes, so in 5 minutes it turns \( 360^\circ \times \frac{5}{60} = 30^\circ \).
Area swept \( = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{12} \times 22 \times 2 \times 14 = \frac{154}{3} \) cm².
Hence, the area swept by the minute hand in 5 minutes is \( \frac{154}{3} \) cm².
In simple words: In 5 minutes the minute hand covers 30 degrees, which is a twelfth of the clock face.

Teacher's Note:
1. The minute hand turns 6 degrees every minute.
2. The area swept is the area of a sector.

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 \( \pi = 3.14 \))
Answer:

Radius \( = 10 \) cm and the angle subtended by the chord at the centre is \( 90^\circ \).
(i) Area of the minor segment \( = \) area of the minor sector \( - \) area of \( \triangle AOB = \frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 \) cm².
(ii) The major sector has the angle \( 360^\circ - 90^\circ = 270^\circ \). Area of the major sector \( = \frac{270^\circ}{360^\circ} \times 3.14 \times 10^2 = 235.5 \) cm². (The same result is obtained as the area of the circle \( 314 \) minus the minor sector \( 78.5 \).)
Hence, the minor segment is 28.5 cm² and the major sector is 235.5 cm².
In simple words: Segment area = sector area minus triangle area; the major sector is three-fourths of the circle.

Teacher's Note:
1. The triangle formed by the two radii and the chord is a right triangle here, so its area is half the product of the radii.
2. Check: 78.5 + 235.5 = 314, the whole circle.

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.
Answer:

(i) Length of the arc \( = \frac{60^\circ}{360^\circ} \times 2\pi r = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = 22 \) cm.
(ii) Area of the sector \( = \frac{60^\circ}{360^\circ} \times \pi r^2 = \frac{1}{6} \times \frac{22}{7} \times 21 \times 21 = 231 \) cm².
(iii) In \( \triangle AOB \), \( OA = OB = 21 \) cm and \( \angle O = 60^\circ \), so \( \angle A = \angle B = 60^\circ \) and the triangle is equilateral. Area of \( \triangle AOB = \frac{\sqrt{3}}{4} \times 21^2 = \frac{441\sqrt{3}}{4} \) cm².
Area of the segment \( = \) area of the sector \( - \) area of \( \triangle AOB = \left(231 - \frac{441\sqrt{3}}{4}\right) \) cm².
Hence, the arc is 22 cm, the sector is 231 cm² and the segment is \( \left(231 - \frac{441\sqrt{3}}{4}\right) \) cm².
In simple words: The 60 degree triangle with equal radii is equilateral, which gives its area quickly.

Teacher's Note:
1. A triangle with two equal sides and a 60 degree angle between them is equilateral.
2. Segment area = sector area minus triangle area.

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 \( \sqrt{3} = 1.73 \))
Answer:

Radius \( r = 15 \) cm and \( \theta = 60^\circ \). Area of the sector \( OAPB = \frac{60^\circ}{360^\circ} \times \pi \times 15^2 = \frac{225}{6}\pi = \frac{225}{6} \times 3.14 = 117.75 \) cm².
\( \triangle AOB \) is equilateral (two equal radii and the angle \( 60^\circ \)), so its area \( = \frac{\sqrt{3}}{4} \times 15^2 = \frac{225 \times 1.73}{4} = 97.31 \) cm².
Area of the minor segment \( = 117.75 - 97.31 = 20.44 \) cm².
Area of the major segment \( = \pi r^2 - \) minor segment \( = 3.14 \times 225 - 20.44 = 706.5 - 20.44 = 686.06 \) cm².
Hence, the minor segment is 20.44 cm² and the major segment is 686.06 cm².
In simple words: Subtract the triangle from the sector for the minor segment, and the minor segment from the circle for the major segment.

Teacher's Note:
1. Use the given values of pi and root 3 only at the final numerical step.
2. Major segment = circle minus minor segment.

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 \( \sqrt{3} = 1.73 \))
Answer:

Draw \( OD \perp AB \). The perpendicular from the centre bisects the chord, so \( AD = DB \), and \( \angle AOD = 60^\circ \), \( \angle OAD = 30^\circ \).
In \( \triangle AOD \): \( \cos 30^\circ = \frac{AD}{12} \Rightarrow AD = 6\sqrt{3} \) cm and \( \sin 30^\circ = \frac{OD}{12} \Rightarrow OD = 6 \) cm. So \( AB = 12\sqrt{3} \) cm.
Area of \( \triangle AOB = \frac{1}{2} \times AB \times OD = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3} = 62.28 \) cm².
Area of the sector \( = \frac{120^\circ}{360^\circ} \times 3.14 \times 12 \times 12 = 150.72 \) cm².
Area of the segment \( = 150.72 - 62.28 = 88.44 \) cm².
Hence, the area of the segment is 88.44 cm².
In simple words: Drop a perpendicular from the centre to the chord to find the triangle's base and height, then subtract the triangle from the sector.

Teacher's Note:
1. The perpendicular from the centre to a chord bisects the chord.
2. Here the triangle is not equilateral, so use trigonometry for its sides.

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. 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 \( \pi = 3.14 \))
Answer:

At a corner of the square the horse can graze a quadrant, a sector of \( 90^\circ \), because the sides of the field form a right angle there.
(i) With \( r = 5 \) m: area \( = \frac{90^\circ}{360^\circ} \times 3.14 \times 5 \times 5 = \frac{1}{4} \times 78.5 = 19.625 \) m².
(ii) With \( r = 10 \) m (still inside the 15 m field): area \( = \frac{90^\circ}{360^\circ} \times 3.14 \times 10 \times 10 = 78.5 \) m².
Increase in grazing area \( = 78.5 - 19.625 = 58.875 \) m².
Hence, the horse grazes 19.625 m² with the 5 m rope and the area increases by 58.875 m².
In simple words: At a corner the grazing region is a quarter of a circle.

Teacher's Note:
1. A corner of a square gives a 90 degree sector.
2. The 10 m rope is shorter than the 15 m side, so the sector still fits in the field.

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. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.
Answer:
Diameter of the circle \( = 35 \) mm, so its radius is \( \frac{35}{2} \) mm.
(i) Circumference of the circle \( = 2\pi r = 2 \times \frac{22}{7} \times \frac{35}{2} = 110 \) mm. Length of the 5 diameters \( = 5 \times 35 = 175 \) mm. Total length of wire \( = 110 + 175 = 285 \) mm.
(ii) Area of the circle \( = \pi r^2 = \frac{22}{7} \times \left(\frac{35}{2}\right)^2 = \frac{1925}{2} \) mm². Area of each sector \( = \frac{1}{10} \times \frac{1925}{2} = \frac{385}{4} \) mm².
Hence, the wire needed is 285 mm and each sector has area \( \frac{385}{4} \) mm² (96.25 mm²).
In simple words: Add the circumference and five diameters; divide the circle's area by 10 for one sector.

Teacher's Note:
1. Five diameters cut the circle into ten equal sectors.
2. Total wire = circumference + 5 diameters.

Question 10. An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Answer:
The 8 equally spaced ribs divide the circle into 8 equal sectors, so the angle of each sector is \( \theta = \frac{360^\circ}{8} = 45^\circ \). The radius is \( r = 45 \) cm.
Area between two consecutive ribs \( = \frac{45^\circ}{360^\circ} \times \frac{22}{7} \times 45 \times 45 = \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22275}{28} \) cm².
Hence, the area is \( \frac{22275}{28} \approx 795.54 \) cm².
In simple words: Each rib gap is one-eighth of the circle.

Teacher's Note:
1. Equal spacing means equal sector angles.
2. The ribs are radii of the circle.

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.
Answer:
Each blade sweeps a sector with \( r = 25 \) cm and \( \theta = 115^\circ \). There are two blades.
Total area cleaned \( = 2 \times \frac{115^\circ}{360^\circ} \times \frac{22}{7} \times 25 \times 25 = \frac{158125}{126} \) cm².
Hence, the total area cleaned at each sweep is \( \frac{158125}{126} \approx 1254.96 \) cm².
In simple words: Two equal sectors are cleaned, so double the sector area.

Teacher's Note:
1. The wipers do not overlap, so the areas simply add.
2. Keep the fraction exact and convert at the end.

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 \( \pi = 3.14 \))
Answer:

The light covers a sector with \( r = 16.5 \) km and \( \theta = 80^\circ \).
Area \( = \frac{80^\circ}{360^\circ} \times 3.14 \times 16.5 \times 16.5 = \frac{2}{9} \times \frac{314}{100} \times \frac{165}{10} \times \frac{165}{10} = 189.97 \) km².
Hence, the area of the sea over which ships are warned is about 189.97 km².
In simple words: The red light covers a sector, so use the sector area formula.

Teacher's Note:
1. The lighthouse is the centre and the range is the radius.
2. Use pi = 3.14 as instructed.

Question 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 Rs. 0.35 per cm². (Use \( \sqrt{3} = 1.7 \))
Answer:

The six equal designs are the segments cut off by the sides of the inscribed regular hexagon. Each segment belongs to a sector of angle \( \theta = \frac{360^\circ}{6} = 60^\circ \), with \( r = 28 \) cm.
Area of each sector \( = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 28 \times 28 = \frac{1}{6} \times 22 \times 4 \times 28 = 410.67 \) cm².
In \( \triangle AOB \), \( OA = OB = 28 \) cm and \( \angle AOB = 60^\circ \), so it is equilateral. Its area \( = \frac{\sqrt{3}}{4} \times 28^2 = \frac{1.7}{4} \times 784 = 333.2 \) cm².
Area of one design \( = 410.67 - 333.2 = 77.47 \) cm², so the area of 6 designs \( = 6 \times 77.47 = 464.82 \) cm².
Cost of making the designs \( = 464.82 \times 0.35 = \) Rs. 162.69.
Hence, the total cost is Rs. 162.69.
In simple words: Each design is a segment: a 60 degree sector minus an equilateral triangle.

Teacher's Note:
1. The sides of a regular hexagon are equal to the radius of its circumscribed circle.
2. Cost = area times rate.

Question 14. Tick the correct solution 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:
Area of a sector \( = \frac{\theta}{360^\circ} \times \pi R^2 \). With \( \theta = p \): area \( = \frac{p}{360} \times \pi R^2 = \frac{p}{360} \times \frac{2}{2} \times \pi R^2 = \frac{p}{720} \times 2\pi R^2 \).
Options (a) and (c) are arc-length type expressions (not areas), and (b) is twice the sector area.
Hence the correct option is (d).
In simple words: Write p/360 times pi R squared, then multiply and divide by 2.

Teacher's Note:
1. Arc length uses 2 pi R, area uses pi R squared.
2. Option (b) would be the area for a 2p degree sector.