NCERT Solutions For Class 10 Math Chapter – 12 Exercise – 12.2

NCERT Solutions For Class 10 Math Chapter – 12 Exercise – 12.2

Q1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60⁰.

Solution:-

Area of the sector = (θ/360⁰) x ∏r²

= 132/7 cm²

Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:-

Area of sector = (90⁰/360⁰) x ∏r² cm²

= (1/4) x (7/2)² ∏ = (49/16) ∏ cm² = 77/8 cm²

Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:-

Radius = 14 cm

Angle rotated by minute hand in 5 minute = 360⁰ x 5/60 = 30⁰

Area of sector = (1/2) x 14² ∏ = 196/12 ∏ cm² = 154/3 cm²

Area swept by the minute hand in 5 minute = 154/3 cm².

Q4. A chord of a circle of radius 10 cm subtends a right angle at the center Find the area of the corresponding (I) Minor segment (ii) Major sector

Solution:-

Radius = 10 cm

Major segment is making 360⁰ – 90⁰ = 270⁰

Area of the sector making angle = 270⁰

= (270⁰/360⁰) x ∏

= (3/4) x 10² ∏

= 75

= 75 x 3.14 cm² = 235.5 cm²

Area of the major segment = 235.5 cm²

Area of ∆AOB = 1/2 X 10 X 10

= 50 cm²

Area of sector = (90⁰/360⁰) x ∏r²

= 25 x 3.14 cm² = 78.5 cm²

Area of the minor segment = Area of the sector making angle 90⁰ – Area of ∆AOB

= 78.5 cm² – 50 cm² = 28.5 cm²

Q5. In a circle of radius 21 cm , an arc subtends an angle of 60⁰ at the centre. Find:

The length of the arc
Area of the sector formed by the arc.
Area of the segment formed by the corresponding chord

Solution:-

Ncert solution class 10 chapter 12-3

Length of the arc AB = θ/360⁰x 2∏r

= 60⁰/360⁰ x 2 x 22/7 x 21

= 22 cm

Angle subtend by the arc = 60⁰

Area of the sector = (60⁰/360⁰) x ∏r² cm²

441/6 x 22/7 cm²

= 231 cm²

Area of equilateral ∆AOB = √3/4 x 21²

= (441√3)/4 cm²

Q6. A chord of a circle of radius 15 cm subtends an angle of 60⁰ at the centre. Find the areas of the corresponding minor and major segments of the circle.

Solution:-

R = 15 cm

Sum of all triangles = 180⁰

∟A + ∟B + ∟C = 180⁰

2∟A = 180⁰ – 60⁰

∟A = 60⁰

Triangle is equilateral as ∟A = ∟B = ∟C = 60⁰

OA = OB = AB = 15 cm

Area of equilateral ∆AOB = √3/4 X (OA)² = √3/4 X 15²

= 97.3 cm²

Angle subtend at the center by minor segment = 60⁰

Area of minor sector = (60⁰ / 360⁰) x∏r² cm²

= 117.75 cm²

Area of the minor segment = Area of Minor sector – Area of equilateral ∆AOB

= 117.75 – 97.3

= 20.4 cm²

Angle made by Major sector = 360⁰ – 60⁰ = 300⁰

Area of the sector = (300⁰/360⁰) x ∏r²

= 588.75 cm²

Area of major segment = Area of Minor sector + Area of equilateral ∆AOB

= 588.75 + 97.3

= 686.05 cm²

Q7. A Chord of a circle of radius 12 cm subtends an angle of 120⁰ at the centre. Find the area of the corresponding segment of the circle.

Solution:-

Cos 30⁰ = AD/OA

√3/2 = AD/12

AD = 6√3 cm.

AB = 2 X AD

= 12√3 cm

Sin 30⁰ = OD/OA

1/2 = OD/12

OD = 6 cm

Area of ∆AOB = 1/2 X Base x Height

= 62.28 cm²

Angle made by minor sector = 120⁰

Area of the sector = (120⁰/360⁰) x ∏r²

= 150.72 cm²

Area of the corresponding Minor segment = Area of the Minor sector – Area of ∆AOB

= 150.72 – 62.28

= 88.44 cm²

Q8. A horse is tied to a peg at one another of a square shaped grass field of side 15 m by means of a 5 m

The area of that part of the field in which the horse can graze.
The increases in the grazing area if the rope were 10 m long instead of 5 m.

Solution:-

Area of circle = ∏r²= 3.14 x 5² = 78.5 m²

Area of that part of the field in which the horse can graze = 1/4 of area of the circle = 78.5/4 = 19.625 m²

Area of circle if the length of rope is increased to 10m = ∏r²

3.14 x 10²

= 314 m²

Area of that part of the field in which the horse can graze = 1/4 of area of the circle

= 314/4

= 78.5 m²

Increase in grazing area = 78.5 m² – 19.625 m² = 58.875 m²

Q9. A broach 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 12.12 Find:

Th total length of the silver wire required
The area of each sector of the broach.

Solution:-

Total length of silver wire required = Circumference of the circle + Length of 5 diameter

2∏r + (5 x 35) mm = (2 x 22/7 x 35/2) + 175 mm

= 110 + 175 mm

= 185 mm

Area of each sector = Total area/Number of sectors

Total area = ∏r²

= 22/7 x (35/2)²

= 1925/2 mm²

Area of each sector = (1925/2) x 1/10 = 385/4 mm²

Q10. 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.

Solution:-

Area between the two consecutive ribs of the umbrella = Total area/Number of ribs

Total Area = ∏r²

= 22/7 x (45)²

= 6364.29/8 cm²

= 795.5 cm²

Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115⁰ Find the total area cleaned at each sweep of the blades.

Solution:-

Area of the sector = (115⁰/360⁰) x ∏r²

= 23/72 x 22/7 x 25²

= 158125/252 cm²

Total area cleaned at each sweep of the blades = 2 x 158125/252 cm²

= 1254.96 cm²

Q12. To warn ships for underwear rocks , a lighthouse spreads a red coloured light over a sector of angle 80⁰ to a distance of 16.5 km. Find the area of the sea over which the ships are warned.

Solution:-

Area of the sector = (80⁰/360⁰) x ∏r² km²

= 2/9 x 3.14 x (16.5)²

= 189.97 km²

Q13. A round table cover has six equal designs as shown in Fig. 12.14 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 .