RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A

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RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A

Question 1:

Radius of the cylinder = 14 m
And its height = 3 m
Radius of cone = 14 m
And its height = 10.5 m
Let l be the slant height

Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)

Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Question 2:

For the cylindrical portion, we have radius = 52.5 m and height = 3 m
For the conical portion, we have radius = 52.5 m
And slant height = 53 m
Area of canvas = 2rh + rl = r(2h + l)

Question 3:

Height of cylinder = 20 cm
And diameter = 7 cm and then radius = 3.5 cm
Total surface area of article
= (lateral surface of cylinder with r = 3.5 cm and h = 20 cm)

Question 4:

Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area

Radius of hemisphere = 4.2 cm
Surface area of two hemispheres

Total surface area = (100.8 + 70.56) π cm²
= 538.56 cm²
= 171.36 π
= 171.36 × 227  cm²
= 538.56 cm²
Further, volume of cylinder = πr²h = 4.2 × 4.2 × 12 π cm²
= 211.68 π cm²
Volume of two hemispheres = 2 × 23 πr³ cu.units
= 43 π  × 4.2 × 4.2 × 4.2
= 98.784 cm³
Volume of wood left = (211.68 – 98.784) π
= 112.896 π cm³
= 112.896 × 227  cm³
= 354.816 cm³

Question 5:
Radius o f cylinder = 2.5 m
Height of cylinder = 21 m
Slant height of cone = 8 m
Radius of cone = 2.5 m
Total surface area of the rocket = (curved surface area of cone + curved surface area of cylinder + area of base)

Question 6:

Height of cone = h = 24 cm
Its radius = 7 cm

Total surface area of toy

Question 7:

Height of cylindrical container h₁ = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container = 
Height of cone r₂ = 12 cm
Diameter = 6 cm
Radius of r₂ = 3 cm

Radius of hemisphere = 3 cm
Volume of hemisphere = 
Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones

Number of cones that can be filled = 10

Question 8:

Diameter of cylindrical gulabjamun = 2.8 cm
Its radius = 1.4 cm
Total height of gulabjamun = AC + CD + DB = 5 cm
1.4 + CD + 1.4 = 5
2.8 + CD = 5
CD = 2.2 cm
Height of cylindrical part h = 2.2 cm
Volume of 1 gulabjamun = Volume of cylindrical part + Volume of two hemispherical parts

Volume of 45 gulabjamuns = 45 × 25.07 cm³
Quantity of syrup = 30% of volume of gulabjamuns
= 0.3 × 45 × 25.07 = 338.46 cm³

Question 9:
Diameter = 7cm, radius = = 3.5 cm
Height of cone = 14.5 cm – 3.5 cm = 11 cm

Total surface area of toy = 

Question 10:
Diameter of cylinder = 24 m
Radius of cylinder = 242 = 12 cm
Height of the cylinder = 11 m
Height of cone = (16 – 11) cm = 5 cm
Slant height of the cone l = 
Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)

Question 11:
Radius of hemisphere = 10.5 cm
Height of cylinder = (14.5 – 10.5) cm = 4 cm
Radius of cylinder = 10.5 cm
Capacity = Volume of cylinder + Volume of hemisphere

Question 12:

Height of cylinder = 6.5 cm
Height of cone = h₂ = (12.8-6.5) cm = 6.3 cm
Radius of cylinder = radius of cone
= radius of hemisphere
= 72 cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere

Question 13:

Radius of each hemispherical end = 282 = 14 cm
Height of each hemispherical part = Its Radius
Height of cylindrical part = (98 – 2 × 14) = 70 cm
Area of surface to be polished = 2(curved surface area of hemisphere) + (curved surface area of cylinder)

Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:

Radius of cylinder r₁ = 5 cm
And height of cylinder h₁ = 9.8 cm
Radius of cone r = 2.1 cm
And height of cone h₂ = 4 cm
Volume of water left in tub = (volume of cylindrical tub – volume of solid)

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm

Volume of cylinder

Volume of cone removed

(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm² = 96 π cm²

Question 16:

Diameter of spherical part of vessel = 21 cm

Question 17:

Height of cylindrical tank = 2.5 m
Its diameter = 12 m, Radius = 6 m
Volume of tank = 
Water is flowing at the rate of 3.6 km/ hr = 3600 m/hr
Diameter of pipe = 25 cm, radius = 0.125 m
Volume of water flowing per hour

Question 18:

Diameter of cylinder = 5 cm
Radius = 2.5 cm
Height of cylinder = 10 cm
Volume of cylinder = πr²h  cu.units = 3.14 × 2.5 × 2.5 × 10  cm³ = 196.25 cm³
Apparent capacity of glass = 196.25
Radius of hemisphere = 2.5 cm
Volume of hemisphere

Actual capacity of glass = ( 196.25 – 32.608 ) cm³ = 163.54 cm³

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