RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Exercise 16.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Surface Area and Volume |

Exercise |
Exercise 16.3 |

Number of Questions Solved |
14 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3

Ex 16.3 Class 10 RBSE Question 1.

The height and radius of base of a cone are 28 cm and 21 cm respectively. Find its (RBSESolutions.com) curved surface area, total surface area and the volume.

Solution :

Given, height of cone (h) = 28 cm.

Radius of base (r) = 21 cm

Exercise 16.3 Class 10 RBSE Question 2.

The volume of a right circular cone is 1232 cm^{3} and its (RBSESolutions.com) height is 24 cm. Find the slant height of the cone.

Solution :

Given, height of cone (h) = 24 cm

Let radius of the cone be r.

Volume of the cone = 1232 cm^{3}

∴ radius of the cone = 7 cm

∵ slant height of cone (l)

Hence, the slant height of the cone = 25 cm.

RBSE Class 10 Maths Chapter 16.3 Question 3.

Diameter and slant height of a (RBSESolutions.com) cone are 14 m and 25 m respectively. Find it total surface area.

Solution :

Given, the base diameter of cone = 14 m

∴ Radius (r) = \(\frac { 14 }{ 5 }\) = 7 m

Slant height (d) = 25 m

Total surface area of the cone = πr(l+r)

= \(\frac { 22 }{ 7 }\) × 7 × (25 + 7)

= \(\frac { 22 }{ 7 }\) × 7 × 32

= 22 × 32

= 744 m^{2}

Hence, total surface area of cone 704 m^{2}

RBSE Solutions For Class 10 Maths Chapter 16.3 Question 4.

The radius of the base of a cone is 14 cm and its slant height is 50 cm. Find the (RBSESolutions.com) curved surface area and total surface area of the cone.

Solution :

Given

The base radius of cone (r) = 14 cm.

Slant height (l) = 50 cm

∴Curved surface area of cone = πrl

= \(\frac { 22 }{ 7 }\) × 14 × 50

= 2200 cm^{2}

And total surface area of cone = πr (l + r)

= \(\frac { 22 }{ 7 }\) × 14 × (50+14)

= 22 × 2 × 64

= 2816 cm^{2}

Hence, curved surface area 2200 cm^{2}

and total surface area = 2816 cm^{2}

16.3 Class 10 RBSE Question 5.

The height of a right cone is 8 cm and its radius of the (RBSESolutions.com) base is 6 cm. Find the volume of the cone.

Solution :

Given,

Height of cone (h) = 8 cm

base radius (r) = 6 cm

Volume of the cone (V) = \(\frac { 1 }{ 3 }\)πr^{2}h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 6 × 6 × 8

= \(\frac { 2112 }{ 7 }\)

= 301.7 cm^{3}

Hence, volume of the cone = 301.7 cm^{3}

RBSE Solutions For Class 10 Maths Chapter 16 Question 6.

The lateral surface area and the slant height of a (RBSESolutions.com) cone are 1884.4 m^{2} and 12 m respectively. Find its base radius.

Solution :

Given,

Slant height of cone (l) = 12 meter

Lateral surface area = 1884.4 m^{2}

Let the radius of base be r.

∴ πrl = 1884.4

\(\frac { 2112 }{ 7 }\) × 12 × r = 1884.4

r = \(\frac { 1884.4\times 7 }{ 22\times 12 }\)

= 50 m(approx)

RBSE Solutions For Class 10 Maths Chapter 16.3 Question 7.

Base area of a right cone is 154 cm^{2}. If its slant height is 25 cm. (RBSESolutions.com) Find the height of the cone.

Solution :

Given,

Slant height of the cone (l) = 25 cm

Base area = 154 cm^{2}

A cone has a circular base,

Let the height of the cone be h,

Hence, the height of the cone = 24 cm.

Class 10 Maths RBSE Solution Chapter 16 Question 8.

Two cones have same diameters of the base. The ratio of (RBSESolutions.com) their slant heights is 5 : 4. If the lateral surface area of the smaller cone is 400 cm^{2}. find the lateral surface area of the larger cone.

Solution :

Given,

Two cones have same base diameters. So their radii will also be equal.

Let r_{1} = r_{2} = r.

Let the slant heights S be l_{1} and l_{1} then according to the question.

\(\frac { { l }_{ 1 } }{ { l }_{ 2 } }\) = \(\frac { 5 }{ 4 }\)

And lateral surface area of smaller cone (πrl_{2}) = 400 cm^{2}

lateral surface area of larger cone = \(\frac { { l }_{ 1 } }{ { l }_{ 2 } }\) × 400

= \(\frac { 4 }{ 5 }\) × 100 = 500 cm^{2}

RBSE Solutions For Class 10 Maths Chapter 16.3 Question 9.

The ratio of slant height and the radius of a cone is 7 : 4. If its lateral (RBSESolutions.com) surface area is 792 cm^{2}, then find its radius.

Solution :

Given, slant height (l) : radius (r) = 7 : 4

Hence, radius of the cone = 12 cm.

Chapter 16 Maths Class 10 RBSE Question 10.

The circumference of the base of conical tent with a (RBSESolutions.com) height 9 m is 44 m. Find the volume of the air inside it.

Solution :

Given,

Height of the cane (h) = 9 m

Circumference of the base = 44 m

∴ 2πr = 44

RBSE Solutions For Class 10 Maths Chapter 16 Miscellaneous Question 11.

A conical vessel with base radius 10 cm. contains some water in it. The water (RBSESolutions.com) level in the vessel is 18 cm high. It is pored into another cylindrical vessel with radius 5 cm. Find the water level in the cylindrical vessel.

Solution :

Base radius of conical vessel (R) = 10 cm

And the height (H) = 18 cm

Volume of conical vessel = \(\frac { 1 }{ 3 }\)πR^{2}H

= \(\frac { 1 }{ 3 }\) × π × (10)^{2} × 18

= π ×100 × 6

= 600 π cm^{3}

Let the water lavel in the cylindrical vessel be h.

Radius (r) = 5 cm.

Now according to the problem

Volume of water in the cylindrical vessel = Volume of water in conical vessel

πr^{2}h = 600 π

or r^{2}h = 600

⇒ (5)^{2}h = 600

⇒ 25 h = 600

h = \(\frac { 600 }{ 25 }\)

= 24 cm

Hence the water level in the cylindrical vessel = 24 cm.

Ch 16 Maths Class 10 RBSE Question 12.

A largest right circular cone ¡s formed by cutting a wooden (RBSESolutions.com) cubical piece of 14 cm side. Find the volume of the cone.

Solution :

Given,

Side of the cube (a) = 14 cm.

The diameter of the largest cube cut off = 14 cm

The radius of ths cone r = \(\frac { 14 }{ 2 }\) = 7 cm

The height of the cone cut off (h) = 14 cm.

∴ Volume of cone = \(\frac { 1 }{ 3 }\)πr^{2}h

\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 14

= \(\frac { 2156 }{ 3 }\)

Hence, the volume of cubical piece cut off = 718.67 cm^{3}.

Class 10 Maths RBSE Solution Chapter 16.3 Question 13.

The base radius and height of a cube are 7 cm and 24 cm respectively. Find it (RBSESolutions.com) slant height, lateral surface area, total surface area and volume.

Solution :

Given,

The base radius of cone (e) = 7 cm

Height (h) = 24 cm

Hence, slant height of cone 25 cm, curved surface area = 550 cm^{2}, total surface area = 704 cm^{2} and volume = 1232 cm^{3}

Class 10 Maths RBSE Solution Chapter 16 Exercise 16.3 Question 14.

Radius and the angle of sector are 12 cm and 120° respectively. By (RBSESolutions.com) making its straight cores a cone is formed. Find its volume.

Solution :

Given,

Radius of the sector OA = OB (r) = 12 cm

Angle of the sector = 120°

We hope the given RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Exercise 16.3, drop a comment below and we will get back to you at the earliest.