In this chapter, we provide RD Sharma Class 10 Ex 16.3 Solutions Chapter 16 Surface Area And Volume for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 16.3 Solutions Chapter 16 Surface Area And Volume pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 16 |

Chapter Name | Surface Area And Volume |

Exercise | 16.3 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Solutions for Class 10 Chapter** **16**** Surface Area And Volume** Ex 16.3 Download PDF

**Surface Area And Volume**Ex 16.3 Download PDF

**Chapter 16: Surface Area And Volume Exercise – 16.3**

**Question: 1**

**A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also find the cost of tin sheet for making the bucket at the rate of Rs 1.20 per **dm^{2}

**Solution:**

Given Diameter to top of bucket = 40 cm

Radius (r_{1}) = 40/2 = 20 cm

Diameter of bottom part of the bucket = 20 cm

Radius (r_{2}) = 30/2 = 10cm

Depth of the bucket (h) = 12 cm

= π/3(20^{2} + 10^{2} + 20 × 10)12

= 8800 cm^{3}

Given that cost of tin sheet used for making bucket per dm^{2} = Rs 1.20

So total cost for 17.87dm^{2} = 1.20 × 17.87 = Rs 21.40. ?

Cost for 17.87dm^{2} = Rs 21.40

**Question: 2**

A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm and height 3 cm. find the area of its whole surface and volume.

**Solution:**

Given base diameter of cone (d_{1}) = 20 cm

Radius (r_{1}) = 20/2 cm = 10 cm

Top diameter of Cone (d_{2}) = 12 cm

Radius (r_{2}) = 12/2 cm = 6 cm

Height of the cone (h) = 3 cm

Volume of the frustum right circular cone

= π/3(10^{2} + 6^{2 }+ 10 × 6)^{3}

= 616 cm^{3}

Let ‘L’ be the slant height of cone

⟹ L = 5 cm

Slant height of cone (L) = 5 cm Total surface area of the cone

= π(10 + 6) × 5 + π × 10^{2} + Pi × 6^{2}

= π(80 + 100 + 36)

= π(216)

= 678.85 cm^{2}

Total surface area of the cone = 678.85 cm^{2}

**Question: 3**

The slant height of the frustum of a cone is 4 cm and perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.

**Solution:**

Given slant height of frustum of cone (l) = 4 cm

Let ratio of the top and bottom circles be r_{1} and r_{2}

Given perimeters of its ends as 18 cm and 6 cm

⟹ 2πr_{1} = 18 cm; 2πr_{2} = 6 cm

⟹ πr_{1}= 9 cm … (a)

πr_{2} = 3 cm … (b)

Curved surface area of frustum cone = π(r_{1} + r_{2})l

= π(r_{1 }+ r_{2})l

= (πr_{1}+πr_{2})l = (9 + 3) × 4 = (12) × 4 = 48 cm^{2}

Curved surface area of the frustum cone = 48 cm

**Question: 4**

**The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.**

**Solution:**

Let the slant height of frustum right circular cone be L

L = 16.1 cm

Slant height of the frustum cone = 20.37 cm

Curved surface area of the frustum cone = π(r_{1} + r_{2})l

= π(7 + 5.25)16.1

Curved surface area of the frustum cone = 619.65 cm^{3}

Volume of the frustum cone

= 1898.56 cm^{3}

Volume of the cone = 1898.56 cm^{3}

Total surface area of the frustum cone

= π(7 + 5.25) × 16.1 + π7^{2 }+ π5.25^{2}

= π(7 + 5.25) × 16.1 + π(7^{2} + 5.25^{2}) = 860.27 cm^{2}

Total surface area of the frustum cone = 860.27 cm^{2}

**Question: 5**

**If the radius of circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm. Find the capacity of the bucket.**

**Solution:**

Given Height of the conical bucket = 45 cm Radii of the 2 circular ends of the conical bucket is 28 cm and 7 cm r_{1} = 28 cm r_{2} = 7 cm

Volume of the conical bucket

= 1/3 π(28^{2} + 7^{2} + 28 × 7)45 = 15435π

Volume = 48510 cm^{3}

**Question: 6**

**If the radii of circular end of a bucket 24 cm high are 5 and 15 cm. Find surface area of the bucket.**

**Solution:**

Given,

Height of the bucket (h) = 24 cm

Radius of the circular ends of the bucket 5 cm and 15 cm r_{1} = 5 cm; r_{2} = 15 cm

Let ‘L’ be the slant height of the bucket

Curved surface area of the bucket

= π(5 + 15)26 + π15^{2} = π(520 + 225) = 745π cm^{2}

Curved surface area of the bucket = 745π cm^{2}

**Question: 7**

**The radii of circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 15 cm. Find the total surface area and volume of frustum.**

**Solution:**

Let slant height of the frustum cone be ‘L’ Given height of frustum cone = 12 cm

Radii of a frustum cone are 12 cm and 3 cm r_{1} = 12 cm; r_{2} = 3 cm

L = 15 cm

Total surface area of cone

= π(12 + 3)15 + π12^{2} + π3^{2}

Total surface area = 378 cm^{2}

Volume of frustum cone

= 1/3 π(12^{2} + 3^{2} + 12 × 3) × 12 = 756π cm^{3}

Volume of the frustum cone = 756 π cm^{3}

**Question: 8**

**A tent consists of a frustum of a cone capped by a cone. If radii of ends of frustum be 13 m and 7 m the height of frustum be 8 m and the slant height of the conical cap be 12 m. Find canvas required for the tent.**

**Solution:**

Given height of frustum (h) = 8 m

Radii of the frustum cone are 13 cm and 7 cm

r_{1} = 13 m; r_{2} = 7

Let ‘L’ be slant height of the frustum cone

Curved surface area of frustum (s_{1}) = π(r_{1} + r_{2}) × L= π(13 + 7) × 10 = 200 π m^{2}

Curved surface area of frustum (s_{1})= 200 π m^{2}

Given slant height of conical cap = 12 m

Base radius of upper cap cone = 7 m

Curved surface area of upper cap cone (s_{2}) = πrL = π × 7 × 12 = 264 m^{2}

Total canvas required for tent (S) = s_{1} + s_{2}

S = 200π + 264 = 892.57m^{2}

Total canvas = 892.57 m^{2}

**Question: 9**

**A bucket is in the form of a frustum of a cone with a capacity of 12308.8 **cm^{3}** of water. Radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making.**

**Solution:**

Radii of top circular ends (r_{1}) = 20 cm

Radii of bottom circular end of bucket (r_{2}) = 12 cm

Let height of bucket be ‘h’ Volume of frustum cone

= 784/3 πh cm^{3} … (a)

Given capacity/ volume of the bucket = 12308.8 cm^{3} … (b) Equating (a) and (b)

⟹ 784/3 πh cm^{3} = 12308.8

⟹ h =15 cm

Height of the bucket (h) = 15 cm

Let ‘L’ be slant height of bucket

⟹ L^{2} = (r_{1} − r_{2})^{2} + h^{2}

⟹ L = 17 cm

Length of the bucket/slant height of the bucket (L) = 17 cm

Curved surface area of bucket

= π(20 + 12)17 + π × 12^{2}

= π(9248 + 144) = 2160.32 cm²

curved surface area = 2160.32 cm²

**Question: 10**

**A bucket made of aluminum sheet is of height 20 cm and its upper and lower ends are of radius 25 cm and 10 cm. Find cost of making if the aluminum sheet costs Rs 70 per 100 **cm^{2}**.**

**Solution:**

Given height of bucket (h) = 20 cm

Upper radius of bucket (r_{1}) = 25 cm

Lower radius of the bucket (r_{2}) = 10 cm

Let ‘L’ be slant height of the bucket

Slant height of bucket (L) = 25 cm

Curved surface area of bucket

= π(25 + 10) × 25 + π × 10^{2} = 3061.5 cm^{2} ?

Curved surface area = 3061.5 cm^{2}

Cost of making bucket per 100 cm^{2} = Rs 70

Cost of making bucket per 3061.5 cm^{2} = 3061.5/100 × 70 = Rs 2143.05

Total cost for 3061.5 cm^{2} = Rs 2143.05

**Question: 11**

**Radii of circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Finds its total surface area.**

**Solution:**

Given slant height of frustum cone = 10 cm

Radii of circular ends of frustum cone are 33 cm and 27 cm r_{1} = 33 cm; r_{2} = 27cm

Total surface area of a solid frustum of cone

= π(33 + 27) × 10 + π33^{2} + π27^{2}

= π(60) × 10 + π33^{2 }+ π27^{2}

= π(600 + 1089 + 729)

= 2418π cm^{2}

Total surface area = 7599.42 cm^{2}

**Question: 12**

**A bucket made up of a metal sheet is in from of a frustum of cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm. Find the volume of the bucket. Also find the cost of the bucket if the cost of metal sheet used id Rs 20 per 100 **cm^{2}**.**

**Solution:**

Given height of frustum cone = 16 cm

Diameter of the lower end of the bucket d_{1} = 16 cm

Lower end radius r_{1} = 16/2 = 8 cm

Upper end radius r_{2} = 340/2 = 20 cm

Let ‘L’ be slant height of frustum of cone

L = 20 cm

Slant height of the frustum cone L = 20 cm

Volume of the frustum cone

=1/3π(8^{2 }+ 20^{2 }+ 8 × 20) × 16

Volume of the frustum cone = 10449.92 cm^{3}

Curved surface area of the frustum cone

Cost of the metal sheet per 100 cm^{2} = Rs 20

Cost of the metal sheet per 624 π cm^{2} = 624π/100 × 20

Cost of the metal sheet = 391.9

**Question: 13**

A solid is in the shape of a frustum of a cone. The diameters of two circular ends are 60 cm and 36 cm and height is 9 cm. find area of its whole surface and volume.

**Solution:**

Given height of the frustum cone = 9 cm

Lower end radius r_{1} = 60/2 = 30cm

Upper end radius r_{2} = 36/2 = 18 cm

Let slant height of the frustum cone be

L = 15 cm

Volume of the frustum cone

=1/3π(30^{2 }+ 18^{2 }+ 30 × 18) × 9 = 5292π cm^{3}

Volume = 5292π cm^{3}

Total surface area of frustum cone

= π(30 + 18) × 15 + π30^{2} + π18^{2}

= π(720 + 900 + 324)

= 1944 Picm^{2}

Total surface area = 1944π cm^{2}

**Question: 14**

**A milk container is made of metal sheet in the shape of frustum cone whose volume is 10459 **cm^{3}**. The radii of its lower and upper circular ends are 8 cm and 20 cm. Find the cost of metal sheet used in making container at a rate of Rs 1.40 per **cm^{2}**.**

**Solution:**

Given,

Lower end radius r_{1} = 8 cm

Upper end radius r_{2} = 20 cm

Let the height of the container be ‘h’

V_{1 }= 1/3π (8^{2} + 20^{2} + 8(20))h cm^{3} …. (1)

Volume of the milk container

V_{1} – V_{2}

1/3π(8^{2} + 20^{2} + 8(20))h cm^{3}

⟹ h = 16 cm

Height of frustum cone (h) = 16 cm

Let slant height of frustum cone be ‘L’

L = 20 cm

Slant height of frustum cone (L) = 20 cm

Total surface area of the frustum cone

= π(20 + 8 × 20 + π20^{2 }+ π8^{2}

= π(560 + 400 + 64)

= 1024π

= 3216.99 cm^{2}

Total surface area of the frustum = 3216.99 cm^{2}

**Question: 15**

**A reservoir in form of frustum of a right circular cone contains **44 × 10^{7}** liters of water which fills it completely. The radii of bottom and top of the reservoir are 50 m and 100 m. Find the depth of water and lateral surface area of the reservoir.**

**Solution:**

Let depth of frustum cone be h Volume of first cone

r_{1} = 50 m; r_{2} = 100

Volume of the reservoir = 44 × 10^{7} liters …. (b)

Equating (a) and (b) 1/3 π(8500)h = 44 × 10^{7} h = 24

**Question: 16**

**A metallic right circular cone 20 cm high and whose vertical angle is **90°** is cut into two parts at the middle point of its axis by a plane parallel to the base. If frustum so obtained be drawn into a wire of diameter **1/16** cm. Find the length of the wire.**

**Solution:**

Let ABC be the cone.

Height of metallic cone AO = 20 cm

Cone is cut into two parts at the middle point of its axis

Hence height of frustum cone AD = 10 cm

Since angle A is right angles.

So each angles B and C = 45°

Angles E and F = 45°

Let radii of top and bottom circle of frustum

Cone be r_{1} and r_{2}

From ΔADE ⟹ DE/AD = Cot 45°

⟹ r_{1}/10 = 1 ⟹ r_{1} = 10 cm

From ΔAOB

⟹ OE/OA = Cot 45°

⟹ r_{2}/20 = 1

⟹ r_{2} = 20 cm

**All Chapter RD Sharma Solutions For Class10 Maths**

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good