In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 19 |

Chapter Name | Volume and Surface Areas of Solids |

Exercise | 19 A |

Category | RS Aggarwal Solutions |

**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 cm^{2}

= 538.56 cm^{2}

= 171.36 π

= 171.36 × 227 cm^{2}

= 538.56 cm^{2}

Further, volume of cylinder = πr^{2}h = 4.2 × 4.2 × 12 π cm^{2}

= 211.68 π cm^{2}

Volume of two hemispheres = 2 × 23 πr^{3} cu.units

= 43 π × 4.2 × 4.2 × 4.2

= 98.784 cm^{3}

Volume of wood left = (211.68 – 98.784) π

= 112.896 π cm^{3}

= 112.896 × 227 cm^{3}

= 354.816 cm^{3}

**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 h_{1} = 15 cm

Diameter of cylindrical container = 12 cm

Volume of container =

Height of cone r_{2} = 12 cm

Diameter = 6 cm

Radius of r_{2} = 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 cm^{3}

Quantity of syrup = 30% of volume of gulabjamuns

= 0.3 × 45 × 25.07 = 338.46 cm^{3}

**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 = h_{2} = (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 r_{1} = 5 cm

And height of cylinder h_{1} = 9.8 cm

Radius of cone r = 2.1 cm

And height of cone h_{2} = 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 cm^{2} = 96 π cm^{2}

**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 = πr^{2}h cu.units = 3.14 × 2.5 × 2.5 × 10 cm^{3 }= 196.25 cm^{3}

Apparent capacity of glass = 196.25

Radius of hemisphere = 2.5 cm

Volume of hemisphere

Actual capacity of glass = ( 196.25 – 32.608 ) cm^{3} = 163.54 cm^{3}

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