NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volume

Mathematics is a subject requiring high knowledge and skills. For one to be good at Mathematics, one must be thorough with the vital concepts and also be comfortable with calculations. Hence, students should know all the techniques associated with the study of Mathematics.

Class 9 Mathematics Chapter 13 is more of a kind of chapter of mensuration covered in lower classes. In order to score well in this chapter, one must be through with all the formulas mentioned in the chapter. The main topics covered in NCERT Class 9 Mathematics Chapter 13 include the total and lateral surface area of cube, cuboid, cylinder, cone and sphere. You will also be acquainted with how to find the volume of cubes, cuboids, cylinders, cones and spheres.

The chapter has been covered in detail in NCERT Solutions for Class 9 Mathematics Chapter 13. Students can find all the formulas compiled in quick notes format, which will facilitate them to quickly grasp the topics in this chapter. Students will also be able to apply these formulas as required in the problems once they refer to our NCERT Solutions.

All the students taking the guidance from Extramarks and following Extramarks resources have seen a lot of changes in their results and have turned out to be great academic performers. Thus, becoming confident and developing a strong personality. You can find all the study material required for your preparation on the Extramarks website.

Key Topics Covered In NCERT Solutions for Class 9 Maths Chapter 13

Mensuration has always been an important part of Geometry as it covers the detailed study of the calculation of areas of 2D and 3D shapes. The chapter surface areas and volumes are another name for mensuration. As the name suggests, the chapter covers all the formulas associated with cube, cuboid, cylinder, cone and sphere in calculating their surface areas and volumes. Thus, it is more of a formula learning-oriented chapter for the students.

Students should read and revise the formulas as much as they can in order to be thorough with the chapter. NCERT Solutions Class 9 Mathematics Chapter 13 has covered all the major points related to this chapter in a very precise manner aiding students to get rid of the fear of multiple formulas included in this chapter. Students can get NCERT Solutions for Class 9 Mathematics Chapter 13 from the Extramarks’ website.

NCERT Solutions for Class 9 Mathematics Chapter 13 requires students to use solutions, reasoning, thinking skills, and facilities in solving sums with a better approach.

Introduction

In this chapter, we will learn about shapes like cuboids, cubes, circular cylinders, circular cones, spheres, etc., and we will learn to find out their surface area and their volume.

Surface area of cuboid and cube

In this part of the chapter, we will find out the surface area of the cuboid and cube. You can use the following formula for the surface area of the cuboid and the cube:

Surface area of cuboid = 2(lb + bh + hl)
Surface area of cube = 6a2.

You can find multiple examples to practise and learn more about this section in our NCERT Solutions for Class 9 Mathematics Chapter 13, available on the Extramarks’ website.

Surface area of a right circular cylinder

In this part of the chapter, we will find out the surface area of the right circular cylinder using the formula:

Total surface area of right circular cylinder = 2πr(r + h)

Surface area of a right circular cone

In this part of the chapter, we will find out the surface area of the right circular cone. The total surface area of the right circular cone is given by:

Total Surface area of the right circular cone = πr(l + r)

Surface area of a sphere and hemisphere.

In this part of the chapter, we will find out the surface area of the sphere and hemisphere. The formulas are as follows:

Surface area of sphere = 4πr2
Surface area of hemisphere = 2πr2.

For more details regarding the chapter and to find solutions to all the exercises given in the chapter in the NCERT Solutions for Class 9 Mathematics Chapter 13.

Volume of cuboid and cube

In this part of the chapter, we will find out the volume of the cuboid and cube. Their volume can be calculated by using the formula:

Volume of cuboid = lbh
Volume of cube = a3

Volume of cylinder

In this part of the chapter, we will find out the volume of the cylinder. The formula for the calculation of volume is

Volume of cylinder = πr2h

Volume of a right circular cone.

In this part of the chapter, we will find out the Volume of a right cylinder cone using the formula:

Volume of a right cylinder cone =⅓* πr2h

Volume of a sphere

In this part of the chapter, we will find out the volume of a sphere and volume of a hemisphere using the formula listed below:

Volume of a sphere = 4/3*πr3
Volume of a hemisphere = 2/3*πr3

Summary

In this chapter, we covered all about the 2D and 3D shapes and the formulas to calculate their areas

Surface area

Surface area of cuboid = 2(lb + bh + hl)
Surface area of cube = 6a2
Total Surface area of right circular cylinder = 2πr(r + h)
Total Surface area of right circular cone = πr(l + r)
Surface area of sphere = 4πr2
Surface area of hemisphere = 2πr2

Volume

- Volume of cuboid = lbh
- Volume of cube = a3

Volume of cylinder = πr2h
Volume of a right cylinder cone =⅓* πr2h
Volume of a sphere = 4/3*πr3
Volume of a hemisphere = 2/3*πr3

The chapter is completely covered in detail in the NCERT Solutions for Class 9 Mathematics Chapter 13 available on the Extramarks’ website.

NCERT Solutions for Class 9 Maths Chapter 13: Exercise & Solutions

Even if one is thorough with all the formulas included in the chapter surface areas and volumes, one must be able to solve the problems accurately till the last step and be able to get the right answers. This mastery comes with a lot of practice. For this reason, we have included solutions to all the exercises given in the NCERT textbook in our NCERT Solutions for Class 9 Mathematics Chapter 13 in a detailed manner. One can also find extra questions related to the chapter on our Extramarks’ website, a trusted source for all the NCERT-related study material.

You can find for exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 13 by referring to the following links:

Chapter 13: Exercise 13.1 Question and answers
Chapter 13: Exercise 13.2 Question and answers
Chapter 13: Exercise 13.3 Question and answers
Chapter 13: Exercise 13.4 Question and answers
Chapter 13: Exercise 13.5 Question and answers
Chapter 13: Exercise 13.6 Question and answers
Chapter 13: Exercise 13.7 Question and answers
Chapter 13: Exercise 13.8Question and answers
Chapter 13: Exercise 13.9Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 13, students can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

NCERT Solutions Class 1
NCERT Solutions Class 2
NCERT Solutions Class 3
NCERT Solutions Class 4
NCERT Solutions Class 5
NCERT Solutions Class 6
NCERT Solutions Class 7
NCERT Solutions Class 8
NCERT Solutions Class 9
NCERT solutions Class 10
NCERT solutions Class 11
NCERT solutions Class 12

NCERT Exemplar for Class 9 Maths

Students must include NCERT Exemplar Class 9 Mathematics in their core study material as the book is designed by the subject matter experts. Thus, it helps students to get an idea of the types of questions they would have to face in their examinations. As a result, students find their examination papers easy after solving questions from the NCERT Exemplar.

The book is designed as per the latest CBSE syllabus. Students can promise themselves success in every examination they face after practising from this book. It has proved to be fruitful for the teachers too in framing their question papers, thus becoming an important resource for the students as well as teachers.

Learning to solve the problems in a well stipulated time is quite necessary to complete the entire examination in a given time frame. Hence, NCERT Exemplar provides quick tips and tricks to solve every sum, thereby making calculations easy. You can get NCERT Exemplar Class 9 Mathematics from the Extramarks’ website today.

Key Features of NCERT Solutions for Class 9 Maths Chapter 13

Students with good calculating skills always excel in Mathematics. Hence, in NCERT Solutions for Class 9 Mathematics Chapter 13, we have covered various aspects which help in making the calculations of the students strong. The key features are as follows:

Multiple Questions are covered from all the topics and sub-topics of the chapter in the NCERT Solutions for Class 9 Mathematics Chapter 13, keeping in mind no concept remains untouched
All sets of questions right from basic to advanced, are covered noting the levels of different kinds of students
All the questions are covered as per the latest CBSE curriculum.
After completing the NCERT Solutions for Class 9 Mathematics Chapter 13, students will be able to distinguish between the areas and volumes of different figures easily and confidently.

Q.1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at he top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m² costs ₹ 20.

Ans.

$\begin{array}{l} Length of the box = 1.5 m \\ Breadth of the box = 1.25 m \\ Height of the box = 65 cm \\ = 0.65 m \\ (i) Surface area of open box \\ = lb+2 (b+l) h \\ = 1.5×1.25+2 (1.25+1.5) ×0.65 \\ = 1.875+3.575 \\ {= 5.45 m}^{2} \\ Thus, the area of the sheet required for making the box is \\ {5.45 m}^{2} . \\ (ii) {Cost of 1m}^{2} sheet = ₹ 20 \\ Cost of sheet = ₹ 20×5.45 = ₹109 \end{array}$

Q.2 The length, breadth and height of a room are 5 m, 4 m and 3 m respectiv y. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m².

Ans.

$\begin{array}{l} The length of a room = 5 m \\ The breadth of a room = 4 m \\ The height of a room = 3 m \\ Surface area of 4 walls and ceiling = lb+2 (b+l) h \\ = 5×4+2 (4+5) ×3 \\ = 20+54 \\ {= 74 m}^{2} \\ {Rate of whitewashing of 1 m}^{2} area = ₹ 7.50 \\ Rate of whitewashing the room = ₹ 7.50×74 \\ = ₹ 555 \\ Thus, the cost of whitewashing the room is ₹ 555. \end{array}$

Q.3 The floor of a rectangular hall has a perimeter 250 m. If the cost of painting e four walls at the rate of Rs 10 per m² is Rs 15000, fin the height of the hall.

Ans.

$\begin{array}{l} The perimeter of a rectangular hall = 250 m \\ {Rate of painting the walls = ₹10 per m}^{2} \\ The cost of painting the 4 walls = ₹15000 \\ Let height of hall = h m \\ Area of 4 walls of the room = 2 (l+b) h \\ = 250 h \\ Cost of painting the four walls = 10×250 h \\ ₹ 15000 = 10×250 h \\ h = \frac{15000}{2500} = 6 m \\ Thus, the height of hall is 6 m. \end{array}$

Q.4 The paint in a certain container is sufficient to paint an area equal to 9.375 m. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Ans.

$\begin{array}{l} {The area to paint by a certain container = 9.375 m}^{2} \\ Dimensions of given brick = 22.5cm×10cm×7.5cm \\ Surface area of a brick = 2 (\begin{array}{l} 22.5cm×10cm+10cm×7.5cm \\ +7.5cm×22.5cm \end{array}) \\ = 2 (225+75+168.25) \\ = 2 (468.75) \\ {= 937.5 cm}^{2} \\ = \frac{937.5}{10000} m^{2} \\ Number of bricks painted in given container = \frac{9.375}{(\frac{937.5}{10000})} \\ = 100 \\ Thus, 100 bricks can be painted out of this container. \end{array}$

Q.5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

Ans.

$\begin{array}{l} (i) Edge of a cubical box = 10 cm \\ Lateral surface area of cubical box \\ = 4 {(10)}^{2} \\ = 400 {cm}^{2} \\ Length of cuboidal box = 12 .5 cm \\ Breadth of cuboidal box = 10 cm \\ Height of cuboidal box = 8 cm \\ Lateral surface area of cuboidal box \\ = 2 (l + b) h \\ = 2 (12 .5 + 10) \times 8 {cm}^{2} \\ = 360 {cm}^{2} \\ cubical box has the greater lateral surface area. \\ Greater area of cubical box = 400 - 360 \\ = 40 {cm}^{2} \\ (ii) Total surface area of cubical box \\ = 6 {(side)}^{2} \\ = 6 {(10)}^{2} \\ = 600 {cm}^{2} \\ Total surface area of cuboidal box \\ = 2 (lb + bh + hl) \\ = 2 (12 .5 \times 10 + 10 \times 8 + 8 \times 12 .5) {cm}^{2} \\ = 2 (125 + 80 + 100) {cm}^{2} \\ = 2 (305) \\ = 610 {cm}^{2} \\ Total surface area of cuboidal box is greater than cubical box. \\ Greater surface area of cubical box \\ = 610 - 600 \\ = 10 {cm}^{2} \end{array}$

Q.6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including ase) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Ans.

The length of greenhouse = 30 cm

The breadth of greenhouse = 25 cm
The height of greenhouse = 25 cm

(i) The area of the glass required
= 2(lb + bh + hl)
= 2(30 x 25 + 25 x 25
+ 25 x 30) cm²
= 4250 cm²

Thus, the area of the glass is 4250 cm².

(ii) Length of required tape = 4(l + b + h)
= 4(30 + 25 + 25)
= 320 cm
Thus, 320 cm tape is needed for all the 12 edges

Q.7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing t ir sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Ans.

Total surface area of bigger box
= 2(lb + bh + hl)
= 2(25×20+ 20×5+ 5×25)
= 2 (725)
= 1450 cm²
Required extra surface area = 5% of 1450 cm²
=72.5 cm²
Total surface area of smaller box
= 2(lb + bh + hl)
= 2(15×12+12×5+ 5×15)
= 2 (315)
= 630 cm²
Required extra surface area = 5% of 630 cm²
=31.5 cm²
Total required cardboard for both types boxes
= 1450 + 72.5 + 630 + 31.5
= 2184 cm²
Total required cardboard for both types 250 each boxes

= 250 x 2184 cm²
= 546000 cm²
Cost of 1000 cm² cardboard = Rs. 4
Cost of required cardboard = Rs. (4/1000) x 546000
= Rs. 2184

Q.8 Parveen wanted to make a temporary shelter for her car, by making a box-like ructure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Ans.

Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?

Q.9 The curved surface area of a right circular cylinder of height 14 cm is 88 cm. Find the diameter of the base of the cylinder

Ans.

$\begin{array}{l} Height of the cylinder = 14 cm \\ {Curved surface area of the cylinder = 88 cm}^{2} \\ {2πrh = 88 cm}^{2} \\ r = \frac{88}{2× (\frac{22}{7}) ×14} \\ = \frac{88}{88} = 1 \\ Diameter of the cylinder = 2r \\ = 2 (1) = 2 cm \end{array}$

Q.10 It is required to make a closed cylindrical tank of height 1 m and base diameter 0 cm from a metal sheet. How many square metres of the sheet are required for the same?

Ans.

$\begin{array}{l} The height of closed cylinderical tank = 1 m \\ The base diameter of cylinderical tank = 140 c m \\ = 1.4 m \\ The base radius of cylinderical tank = \frac{1.4}{2} \\ = 0.7 m \\ Required sheet for cylinder = 2 \times \frac{22}{7} \times 0.7 \times 1 + 2 \times \frac{22}{7} {(0.7)}^{2} \\ = 4.4 m^{2} + 3.08 \\ = 7.48 m^{2} \\ T hus, 7 {.48 m}^{2} sheet is required for closed cylinderical tank . \end{array}$

Q.11 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the o er diameter being 4.4 cm (figure shown below) . Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Ans.

$\begin{array}{l} Length of metal pipe = 77 cm \\ Inner diameter of metal pipe = 4 cm \\ Outer diameter of metal pipe = 4 .4 cm \\ (i) Inner surface area of metal pipe \\ = π \times inner diameter \times h \\ = \frac{22}{7} \times 4 \times 77 \\ = 968 {cm}^{2} \\ (ii) Outer surface area of metal pipe \\ = π \times Outer diameter \times h \\ = \frac{22}{7} \times 4 .4 \times 77 \\ = 1064 .8 {cm}^{2} \end{array}$

$\begin{array}{l} (iii) Tital surface area = Outer surface area \\ +Inner surface area + 2π ({r_{1}}^{2} - {r_{2}}^{2}) \\ =1064.8+968+2× \frac{22}{7} \{{(2.2)}^{2} {-2}^{2}\} [∵ r_{1} = \frac{4.4}{2} {= 2.2, r}_{2} = \frac{4}{2} = 2] \\ = 2032.8+5.28 \\ {= 2038.08 cm}^{2} \\ {Thus, the total surface area of metal pipe is 2038.08 cm}^{2} . \end{array}$

Q.12 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete volutions to move once over to level a playground. Find the area of the playground in m².

Ans.

$\begin{array}{l} The diameter of a roller = 84 cm \\ The radius of a roller = \frac{84}{2} \\ = 42 cm \\ The length of roller = 120 cm \\ Area of playground covered in one revolution \\ = curved surface area of roller \\ = 2 \times \frac{22}{7} \times 42 \times 120 \\ = 31680 {cm}^{2} \\ Area of playground covered in 500 revolutions \\ = 500 \times 31680 {cm}^{2} \\ = 15840000 {cm}^{2} \\ = \frac{15840000}{10000} m^{2} [∵ 1 {cm}^{2} = \frac{1}{10000} m^{2}] \\ = 1584 m^{2} \\ Thus, the area of plalyground is 1584 m^{2} . \end{array}$

Q.13 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of p nting the curved surface of the pillar at the rate of Rs 12.50 per m².

Ans.

$\begin{array}{l} Diameter of a cylindrical pillar = 50 cm \\ = \frac{50}{100} m \\ = \frac{1}{2} m \\ Radius of a cylindrical pillar = \frac{1}{4} m \\ Height of a cylindrical pillar = 3.5 m \\ Curved surface area of pillar = 2πrh \\ = 2× \frac{22}{7} × \frac{1}{4} ×3.5 \\ = \frac{11}{2} m^{2} \\ {Cost of painting of 1m}^{2} area = Rs.12.50 \\ Cost of painting of \frac{11}{2} m^{2} area = Rs.12.50× \frac{11}{2} \\ = Rs.68.75 \\ Therefore, the cost of painting the Curved surface area of the \\ pillar is Rs 68.75. \end{array}$

Q.14 Curved surface area of a right circular cylinder is 4.4 m². If the ra us of the base of the cylinder is 0.7 m, find its height.

Ans.

$\begin{array}{l} {Curved surface area of right circular cylinder = 4.4 m}^{2} \\ Radius of the base of right circular cylinder = 0.7 m \\ {Then, 2πrh = 4.4 m}^{2} \\ 2× \frac{22}{7} {×0.7×h = 4.4 m}^{2} \\ h = \frac{4 .4}{4 .4} m \\ = 1 m \\ Thus, the height of right circular cylinder is 1 m. \end{array}$

Q.15 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find</str g>
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m².

Ans.

$\begin{array}{l} The inner radius of a circular well = \frac{3.5}{2} m \\ The depth of a circular well = 10 m \\ (i) Inner curved surface area of well = 2πrh \\ = 2× \frac{22}{7} × \frac{3.5}{2} ×10 \\ {= 110 m}^{2} \\ (ii) {Cost of plastering 1 m}^{2} of well = ₹ 40 \\ {Cost of plastering 110 m}^{2} of well = ₹ 40×110 \\ = ₹ 4400 \\ Thus, the cost of plastering the inner curved surface of the \\ {well at the rate of ₹ 40 per m}^{2} is ₹4400. \end{array}$

Q.16 In a hot water heating system, there is a cylindrical pipe of length 28 m and dia ter 5 cm. Find the total radiating surface in the system.

Ans.

$\begin{array}{l} Length of hot water heating system = 28 m \\ Diameter of hot water heating system = 5 cm \\ Radius of hot water heating system = \frac{5}{100} m \\ Total radiating surface of the system = 2 πrh \\ = 2 \times \frac{22}{7} \times \frac{5}{100} \times 28 \\ = 8 .8 m^{2} \\ {Thus, the total radiating surface area of the system is 8.8 m}^{2} . \end{array}$

Q.17 Find
( the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if (1/12) of the steel actually used was wasted in making the tank.

Ans.

$\begin{array}{l} The diameter of cylinderical petrol tank = 4.2 m \\ The radius of cylinderical petrol tank = 2.1 m \\ The height of The cylinderical petrol tank = 4.5 m \\ (i) The curved surface area of petrol tank = π×diameter×h \\ = \frac{22}{7} ×4.2×4.5 \\ {= 59.4 m}^{2} \\ Thus, the lateral or curved surface area of a closed cylindrical \\ {petrol storage tank 59.4 m}^{2} . \\ (ii) Total surface area of closed cylinderical tank \\ = 2πr (h+r) \\ = 2× \frac{22}{7} ×2.1 (4.5+2.1) \\ = 44×0.3×6.6 \\ {= 87.12 m}^{2} \\ {Let actually used steel in making tank = x m}^{2} \\ Wasted steel in making cylinderical tank = \frac{1}{12} of x \\ Used steel in making cylinderical tank = x - \frac{x}{12} \\ {87.12 m}^{2} = \frac{11x}{12} \\ x = \frac{12}{11} ×87.12 \\ {= 95.04 m}^{2} \\ {Thus, actually used steel in making cylinderical tank is 95.04 m}^{2} . \end{array}$

Q.18 In Figure, you see the frame of lampshade. It is to be covered with a decorativ cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Ans.

$\begin{array}{l} Diameter of the base of the frame = 20 cm \\ Radius of the base of the frame = 10 cm \\ Height of the frame = 30 cm \\ Length of margin = 2.5 cm \\ Height of the frame with margin = 2.5+30+2.5 \\ = 35 cm \\ Cloth required for covering the lampshade \\ = 2πrh \\ = 2× \frac{22}{7} ×10×35 \\ {= 2200 cm}^{2} \\ {Thus, 2200 cm}^{2} cloth is required for covering the lampshade. \end{array}$

Q.19 The students of a Vidyalaya were asked to participate in a competition for makin and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Ans.

$\begin{array}{l} Radius of penholder = 3 cm \\ Height of penholder = 10.5 cm \\ {Total surface area of a penholder = 2πrh+πr}^{2} \\ = πr (2h+r) \\ = \frac{22}{7} ×3 (2×10.5+3) \\ = \frac{66}{7} {×24 cm}^{2} \\ ∴ Cardboard required for one penholder = \frac{66}{7} {×24 cm}^{2} \\ Cardboard required for 35 penholders = \frac{66}{7} {×24 ×35 cm}^{2} \\ {= 7920 cm}^{2} \\ {Thus, the cardboard required for 35 penholders for competition was 7920 cm}^{2} . \end{array}$

Q.20 Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Ans.

Curved surface area of con

e = (22/7) x (10.5/2) x 10
= 165 cm²
Thus, the curved surface area of cone is 165 cm².

Q.21 Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Ans.

Diameter of the base of cone = 24 m
Radius of the base of cone = 12 m
Slant height of the cone = 21 m

Total surface area of cone =

$π$

r(l + r)
= (22/7) x 12(21 + 12)
= (22/7) x 12 x 33

= 1244.57 cm²
Thus, total surface area of the cone is 1244.57 cm².

Q.22 A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.

Ans.

$\begin{array}{l} Height of conical tent = 10 m \\ Radius of conical tent = 24 m \\ (i) Slant height of the conical tent = \sqrt{{(10)}^{2} + {(24)}^{2}} \\ = \sqrt{100 + 576} \\ = \sqrt{676} \\ = 26 m \\ (i i) Curved surface area of tent = π r l \\ = \frac{22}{7} \times 24 \times 26 m^{2} \\ ∵ {The cost of 1m}^{2} canvas = R s .70 \\ ∴ The cost of the canvas required to make the tent \\ = R s .70 \times \frac{22}{7} \times 24 \times 26 \\ = R s . 137280 \\ Thus,the cost of the canvas required to make the tent is \\ R s . 137280. \end{array}$

Q.23 The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m².

Ans.

$\begin{array}{l} The slant height of conical tomb = 25 m \\ The base diameter of conical tomb = 14 m \\ The base radius of conical tomb = 7 m \\ Curved surface area of conical tomb = π r l \\ = \frac{22}{7} \times 7 \times 25 \\ = 550 m^{2} \\ Rate {of white-washing 100 m}^{2} area = R s .210 \\ {Rate of white-washing 1 m}^{2} area = R s . \frac{210}{100} \\ {Rate of white-washing 550 m}^{2} area = R s . \frac{210}{100} \times 550 \\ = R s .1155 \\ Thus, the cost of white-washing the conical tomb is Rs . 1155 . \end{array}$

Q.24 A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Ans.

$\begin{array}{l} Radius of joker ’ s cap = 7 cm \\ Height of joker ’ s cap = 24 cm \\ Slant height of joker’s cap = \sqrt{7^{2} + 24^{2}} \\ = \sqrt{49 + 576} \\ = \sqrt{625} \\ = 25 c m \\ Curved surface area of cap = π r l \\ = \frac{22}{7} \times 7 \times 25 \\ = 550 c m^{2} \\ Area of the sheet required for 10 caps \\ = 10 \times 550 c m^{2} \\ = 5500 c m^{2} \\ Thus, {the area of required sheet for 10 caps is 5500 cm}^{2} . \end{array}$

Q.25

$\begin{array}{l} A bus stop is barricaded from the remaining part of the road, \\ by using 50 hollow cones made of recycled cardboard . Each \\ cone has a base diameter of 40 cm and height 1 m . If the \\ outer side of each of the cones is to be painted and the cost \\ {of painting is Rs 12 per m}^{2}, what will be the cost of painting \\ all these cones? (Use π = 3 .14 and take \sqrt{1.04} = 1.02) \end{array}$

Ans.

$\begin{array}{l} Diameter of hollow cone = 40 cm \\ Radius of hollow cone = 20 cm \\ = 0.2 m \\ Height of hollow cone = 1 m \\ Slant height of hollow cone = \sqrt{1^{2} + {(0.2)}^{2}} \\ = \sqrt{1 + 0.04} \\ = \sqrt{1.04} m \\ = 1.02 m (approx .) \\ C .S .A . of hollow cone = π r l \\ = 3.14 \times 0.2 \times 1.02 \\ = 0.64056 m^{2} \\ C .S .A . of hollow 50 cones = 50 \times 0.64056 m^{2} \\ = 32.028 m^{2} \\ Cost of painting {of 1 m}^{2} area = R s .12 \\ Cost of painting of 50 cones = R s .12 \times 32.028 \\ = R s . 384.336 \\ = R s . 384.34 (approx) \\ Thus, the cost of painting 50 cones is Rs . 384.34 . \end{array}$

Q.26 Find the surface area of a sphere of radius:

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Ans.

$\begin{array}{l} Surface area of a sphere = 4 π r^{2} \\ (i) Surface area of a sphere = 4 π {(1 0. 5)}^{2} c m^{2} \\ = 4 \times \frac{22}{7} \times 10.5 \times 10.5 c m^{2} \\ = 4 \times 22 \times 1.5 \times 10.5 c m^{2} \\ = 1386 c m^{2} \\ (i i) S u r f a c e area of a sphere = 4 π {(5. 6)}^{2} c m^{2} \\ = 4 \times \frac{22}{7} \times 5. 6 \times 5. 6 c m^{2} \\ = 4 \times 22 \times 0.8 \times 5. 6 c m^{2} \\ = 394.24 c m^{2} \\ (iii) Surface area of a sphere = 4 π {(14)}^{2} c m^{2} \\ = 4 \times \frac{22}{7} \times 14 \times 14 c m^{2} \\ = 4 \times 22 \times 2 \times 14 c m^{2} \\ = 2464 c m^{2} \end{array}$

Q.27 Find the surface area of a sphere of diameter:

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Ans.

$\begin{array}{l} \begin{array}{l} Surface area of a sphere = 4 {πr}^{2} \\ (i) Diameter of sphere = 14 cm \end{array} \\ \begin{array}{l} Radius of sphere = 7 cm \\ Surface area of a sphere = 4 π {(7)}^{2} \end{array} \\ \begin{array}{l} = 4 \times \frac{22}{7} \times 7 \times 7 \\ = 616 {cm}^{2} \end{array} \\ \begin{array}{l} (ii) Diameter of sphere = 21 cm \\ Radius of sphere = \frac{21}{2} cm \end{array} \\ \begin{array}{l} Surface area of a sphere = 4 π {(\frac{21}{2})}^{2} \\ = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \end{array} \\ = 1386 {cm}^{2} \\ \begin{array}{l} (iii) Diameter of sphere = 3 .5 m \\ Radius of sphere = \frac{3 .5}{2} m \end{array} \\ Surface area of a sphere = 4 π (\frac{3 .5}{2}) \\ \begin{array}{l} = 4 \times \frac{22}{7} \times \frac{3 .5}{2} \times \frac{3 .5}{2} \\ = 38 .5 m^{2} \end{array} \end{array}$

Q.28 Find the total surface area of a hemisphere of radius 10 cm. (Use

$π$

= 3.14)

Ans.

$\begin{array}{l} {Total Surface area of a hemisphere = 3πr}^{2} \\ = 3π {(10)}^{2} {cm}^{2} \\ {= 3×3.14×10×10 cm}^{2} \\ {= 942 cm}^{2} \end{array}$

Q.29 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans.

$\begin{array}{l} Surface area of a sphere = 4 π r^{2} \\ When radius of balloon is 7 cm . \\ Surface area of ballon = 4 π {(7)}^{2} \\ = 4 \times \frac{22}{7} \times 7 \times 7 \\ = 616 c m^{2} \\ When radius of balloon is 14 cm . \\ Surface area of ballon = 4 π {(14)}^{2} \\ = 4 \times \frac{22}{7} \times 14 \times 14 \\ = 2464 c m^{2} \\ The ratio of surface areas of balloon in the two cases \\ = \frac{616 c m^{2}}{2464 c m^{2}} \\ = \frac{1}{4} \\ Thus, the ratio of surface areas of the balloon in the \\ two cases is 1:4 . \end{array}$

Q.30 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm².

Ans.

$\begin{array}{l} The inner radius of bowl = \frac{10.5}{2} c m \\ Inner surface area of bowl = 2 π r^{2} \\ = 2 \times \frac{22}{7} \times \frac{10.5}{2} \times \frac{10.5}{2} \\ = 173.25 c m^{2} \\ {Cost of tin-plating 100 cm}^{2} = R s . 16 \\ {Cost of tin-plating 1 cm}^{2} = R s . \frac{16}{100} \\ Cost of tin-plating bowl = R s . \frac{16}{100} \times 173.25 \\ = R s . 27.72 \\ Thus, the cost of tin-platting the inner surface area of bowl is \\ Rs . 27 .72 . \end{array}$

Q.31 Find the radius of a sphere whose surface area is 154 cm².

Ans.

$\begin{array}{l} Let radius of sphere be r cm . \\ Surface area of sphere = 154 c m^{2} \\ 4 π r^{2} = 154 c m^{2} \\ 4 \times \frac{22}{7} \times r^{2} = 154 c m^{2} \\ r^{2} = \frac{154 \times 7}{4 \times 22} c m^{2} \\ = \frac{7 \times 7}{4} c m^{2} \\ r = \frac{7}{2} c m \\ Thus, the radius of sphere is 3 .5 cm . \end{array}$

Q.32 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Ans.

$\begin{array}{l} Surface area of a sphere = 4 π r^{2} \\ Let radius of the earth be R . \\ Let radius of the moon be \frac{R}{4} . [\begin{array}{l} Since, diameter of the moon is \\ \frac{1}{4} of diameter of the earth . \end{array}] \\ Surface area of the earth = 4 π R^{2} \end{array}$ $\begin{array}{l} Surface area of the moon = 4 π {(\frac{R}{4})}^{2} \\ T h e ratio of surface areas of the moon and the earth \\ = \frac{4 π {(\frac{R}{4})}^{2}}{4 π R^{2}} \\ = \frac{1}{16} \\ Thus, the ratio of surface areas of the moon and \\ the earth is 1:16 . \end{array}$

Q.33 A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans.

Inner radius of hemispherical bowl = 5 cm
Thickness of hemispherical bowl = 0.25 cm

Outer radius of hemispherical bowl = 5 + 0.25
= 5.25 cm
So, outer curved surface area of bowl = 2

$π$

r²
=2(22/7)( 5.25)²
= 173.25 cm².
Thus, the outer curved surface area of the bowl is 173.25 cm².

Q.34 A right circular cylinder just encloses a sphere of radius r (see Figure). Find

(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Ans.

$\begin{array}{l} Radius of sphere is r . \\ (i) Surface area of sphere = 4 {πr}^{2} \\ (ii) Height of cylinder = Diameter of sphere \\ = 2 r \\ Radius (R) of cylinder = r \\ Curved surface area of cylinder \\ = 2 πRh \\ = 2 πr (2 r) \\ = 4 {πr}^{2} \\ (iii) Ratio of the surface areas of sphere and cylinder \\ = \frac{4 {πr}^{2}}{4 {πr}^{2}} \\ = 1 \\ Thus, the ratio of the surface areas of sphere and cylinder is 1:1 . \end{array}$

Q.35 A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

Ans.

Volume of a matchbox = 4 cm × 2.5 cm × 1.5 cm

= 15 cm³

Volume of a packet = Volume of 12 matchboxes
= 12 x 15 cm³
= 180 cm³

Thus, the volume of a packet containing 12 matchboxes is 180 cm³.

Q.36 A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m³ = 1000 l)

Ans.

Volume of cuboidal water tank = 6 x 5 x 4.5 m³
= 135 m³
Since, 1 m³ = 1000 litres
So, volume of water in tank = 135 x 1000 litres
= 135000 litres

Q.37 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Ans.

$\begin{array}{l} {Volume of a liquid = 380 m}^{3} \\ Length of cuboidal vessel = 10 m \\ Breadth of cuboidal vessel = 8 m \\ Let height of vessel be h m . \\ Then, \\ volume of cuboidal vessel = l \times b \times h \\ {380 m}^{3} = 10 \times 8 \times h \\ h = \frac{380}{80} \\ = 4.75 m \\ Thus, cuboidal vessel should be 4 .75 m high . \end{array}$

Q.38 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m³.

Ans.

Volume of cuboidal pit = 8 x 6 x 3
= 144 m³
Cost of digging 1 m³ pit = Rs. 30
Then, the cost of digging cuboidal pit
= Rs. 30 x 144
= Rs. 4320
Thus, the cost of digging a cuboidal pit is Rs. 4320.

Q.39 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Ans.

Capacity of cuboidal tank = 50,000 litres
Length of cuboidal tank = 2.5 m
Depth of cuboidal tank = 10 m
Let breadth of cuboidal tank = p m
Since, 1000 litres = 1 m³
So, 50,000 litres = 50 m³
Then, Volume of tank = lbh
50 m³= 2.5 x p x 10
p = 2 m

Thus, the breadth of the tank is 2 m.

Q.40 A village, having a population of 4000, approximately requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?

Ans.

Volume of water tank = 20 m × 15 m × 6 m
= 1800 m³
= 1800000 litres
[Since, 1 m³ = 1000 litres]
Water require for a person = 150 litres
Water require for 4000 persons
= 150 x 4000 litres
= 600000 litres
Number of days to last the water of this tank
= (1800000/600000) days
= 3 days

Thus, the water of tank will last in 3 days in a village of 4000 people.

Q.41 A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

Ans.

$\begin{array}{l} Number of wooden crates in godown \\ = \frac{Volume of godown}{Volume of wooden crates} \\ = \frac{40 m \times 25 m \times 15 m}{1 . 5 m \times 1 . 25 m \times 0. 5 m} \\ = \frac{400 m \times 2500 m \times 150 m}{15 m \times 125 m \times 5 m} \\ = 80 \times 20 \times 10 \\ = 16000 \\ Thus, 16000 wooden crates can be stored in the godown . \end{array}$

Q.42 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Ans.

$\begin{matrix} Side of solid cube = 12 cm \\ Number of cubes of equal volume = 8 \\ Side of each cube of equal volume = \frac{Volume of solid cube}{8} \\ = \frac{12 \times 12 \times 12}{8} \\ = 216 c m^{3} \\ Side of each cube of equal volume = \sqrt[3]{216} \\ = 6 c m \\ Ratio of sufaces area of both cubes = \frac{4 {(12)}^{2}}{4 {(6)}^{2}} \\ = \frac{12 \times 12}{6 \times 6} \\ = 4 : 1 \end{matrix}$ $Thus, the required ratio of surface areas of both cubes is 4:1 .$

Q.43 A part of a river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Ans.

$\begin{array}{l} Speed of river = 2 km per hour \\ = (2000/60) m per minute \\ = \frac{100}{3} m per min \\ Then, \\ Volume of flowing water in one minute \\ = (\frac{100}{3}) (3) (40) \\ {= 4000 m}^{3} \\ {Thus, 4000 m}^{3} water will fall into the sea in a minute. \end{array}$

Q.44 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm³=1 l )

Ans.

$\begin{array}{l} Circumference of the base of a cylindrical vessel \\ = 132 cm \\ 2 πr = 132 cm \\ \Rightarrow 2 \times \frac{22}{7} \times r = 132 cm \\ r = \frac{132 \times 7}{44} \\ r = 21 cm \\ Height of cylinderical vessel = 25 cm \\ Volume of water in the vessel = {πr}^{2} h \\ = \frac{22}{7} \times {(21)}^{2} \times 25 \\ = 34650 {cm}^{3} \\ = \frac{34650}{1000} litres \\ [∵ 1 {cm}^{3} = \frac{1}{1000} litres] \\ = 34 .65 litres \\ Thus, the cyliderical vessel can hold 34.65 litres water . \end{array}$

Q.45 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Ans.

$\begin{array}{l} The inner diameter of a cylinderical wooden pipe \\ = 24 cm \\ The inner radius of a cylinderical wooden pipe \\ = 12 cm \\ The outer diameter of a cylinderical wooden pipe \\ = 28 cm \\ The outer radius of a cylinderical wooden pipe \\ = 14 cm \\ The length of pipe = 35 cm \\ {Mass of 1 cm}^{3} wood = 0 .6 gm \\ Volume of wood in cylinderical wooden pipe \\ = \frac{22}{7} {{(14)}^{2} - {(12)}^{2}} \times 35 \\ = \frac{22}{7} \times 2 \times 26 \times 35 \\ = 5720 {cm}^{3} \\ Mass of wood in cylinderical wooden pipe \\ = 0 .6 \times 5720 gm \\ = 3432 gm \\ = 3 .432 kg \\ Thus, the mass of wood in cylinderical wooden pipe is 3.432 kg. \end{array}$

Q.46 A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Ans.

$\begin{array}{l} (i) Capacity of cuboidal tin can = 5 \times 4 \times 15 \\ = 300 {cm}^{3} \\ (ii) Capacity of plastic cylinder = {πr}^{2} h \\ = \frac{22}{7} \times {(\frac{7}{2})}^{2} \times 10 {cm}^{3} \\ = 385 {cm}^{3} \\ In this way we see that capacity of plastic container is more \\ than tin container. \\ More capacity of cylinderical container than tin container \\ = 385 - 300 \\ = 85 {cm}^{3} \end{array}$

Q.47 It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m², find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.

Ans.

$\begin{array}{l} Depth of cylinder = 10 m \\ Cost of painting the inner curved surface area of cylinder \\ = ₹ 2200 \\ {Rate of painting of 1 m}^{2} = ₹ 20 \end{array}$ $\begin{array}{l} (i)Inner curved surface area of cylinder \\ = \frac{Cost of painting}{Rate of painting} \\ = \frac{2200}{20} \\ = 110 m^{2} \\ (ii) ∵ 2 πrh = 110 m^{2} \\ 2 \times \frac{22}{7} \times r \times 10 = 110 \\ \Rightarrow r = \frac{110 \times 7}{44 \times 10} \\ = 1 .75 m \\ Thus, the radius of the base of cylinder is 1.75 m. \\ (iii) Capacity of vessel = {πr}^{2} h \\ = \frac{22}{7} \times 1 .75 \times 1 .75 \times 10 \\ = 22 \times 0 .25 \times 17 .5 \\ = 96 .25 m^{3} \\ = 96 .25 kl [∵ 1 m^{3} = 1 kl] \end{array}$

Q.48 The volume capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Ans.

$\begin{array}{l} Height of a closed cylinder = 1 m \\ The capacity of a closed cylinder = 15.4 l i t r e s \\ π r^{2} h = \frac{15.4}{1000} m^{3} [1 l i t r e = \frac{1}{1000} m^{3}] \\ \frac{22}{7} \times r^{2} \times 1 = \frac{15.4}{1000} m^{3} \\ r^{2} = \frac{15.4}{1000} \times \frac{7}{22} \\ = \frac{7 \times 7}{10000} \\ \Rightarrow r = \frac{7}{100} = 0.07 m \\ Required metal sheet for closed cylinder \\ = 2 π r (h + r) \\ = 2 \times \frac{22}{7} \times 0.07 (1 + 0.07) \\ = 44 \times 0.01 \times 1.07 \\ = 0.4708 m^{2} \\ Thus, 0.4708 m^{2} of metal sheet would be needed to make \\ closed cylinder . \end{array}$

Q.49 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Ans.

$\begin{matrix} Diameter of the pencil = 7 mm \\ Radius (r_{1}) of the pencil = \frac{7}{2} mm \\ Diameter of the graphite = 1 mm \\ Radius (r_{2}) of the graphite = \frac{1}{2} mm \\ The length of the pencil = 14 cm \\ = 140 mm \\ Volume of the wood = π (r_{1}^{2} - r_{2}^{2}) h \\ = \frac{22}{7} {{(\frac{7}{2})}^{2} - {(\frac{1}{2})}^{2}} \times 140 \\ = 22 (\frac{7}{2} + \frac{1}{2}) (\frac{7}{2} - \frac{1}{2}) \times 20 \\ = 44 \times 4 \times 30 \\ = 5280 m m^{3} \\ = \frac{5280}{1000} c m^{3} [∵ 1 c m = 10 m m] \\ = 5.28 c m^{3} \end{matrix}$

Thus, the volume of wood in the pencil is 5.28 cm³.

$\begin{matrix} V o l u m e of graphite = π r_{2}^{2} h \\ = \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140 \\ = 11 \times 10 \\ = 110 m m^{3} \\ = \frac{110}{1000} c m^{3} \\ = 0.11 c m^{3} \end{matrix}$

Thus, the volume of graphite in the pencil is 0.11 cm³.

Q.50 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare (in litres) daily to serve 250 patients?

Ans.

$\begin{matrix} The diameter of bowl = 7 c m \\ The radius of bowl = \frac{7}{2} c m \\ Height of soup in the bowl = 4 c m \\ Volume of soup in bowl = π r^{2} h \\ = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4 \\ = 154 c m^{3} \\ Volume of soup required for one patient = 154 c m^{3} \\ Volume of soup required for 250 patients = 250 \times 154 c m^{3} \\ = \frac{250 \times 154}{1000} l i t r e s \\ [∵ 1 c m^{3} = \frac{1}{1000} l i t r e s] \\ = 38.5 l i t r e s \end{matrix}$

Thus, 38.5 litres soup the hospital has to prepare daily to serve 250 patients.

Q.51 Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

Ans.

$\begin{array}{l} (i) Volume o f cone = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 6 \times 6 \times 7 \\ = 264 c m^{3} \\ Thus, {the volume of right circular cone is 264 cm}^{3} . \\ (ii) Volume of cone = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12 \\ = 154 c m^{3} \\ Thus, {the volume of right circular cone is 154 cm}^{3} . \end{array}$

Q.52 Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

Ans.

$\begin{array}{l} (i) Radius of cone = 7 cm, Slant height of cone = 25 cm \\ Height of cone (h) = \sqrt{l^{2} - r^{2}} \\ = \sqrt{25^{2} - 7^{2}} \\ = \sqrt{625 - 49} \\ = 24 c m \\ Capacity of vessel = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \\ = 22 \times 7 \times 8 \\ = 1232 c m^{3} \\ = \frac{1232}{1000} l i t r e s [∵ 1 c m^{3} = \frac{1}{1000} l i t r e s] \\ = 1.232 l i t r e s \\ (ii) Height of cone = 12 cm, Slant height of cone = 13 cm \\ Radius of cone (r) = \sqrt{l^{2} - h^{2}} \\ = \sqrt{13^{2} - 12^{2}} \\ = \sqrt{169 - 144} \\ = 5 c m \\ Capacity of vessel = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 12 \\ = \frac{22}{7} \times 25 \times 4 \\ = \frac{2200}{7} c m^{3} \\ = \frac{2200}{7 \times 1000} l i t r e s [∵ 1 c m^{3} = \frac{1}{1000} l i t r e s] \\ = \frac{11}{35} l i t r e s \end{array}$

Q.53 A conical pit of top diameter 3.5 m is 12 m deep of height. What is its capacity in kilolitres?

Ans.

$\begin{array}{l} Diameter of conical pit = 3.5 m \\ Radius of conical pit = \frac{3.5}{2} m \\ Depth of conical pit = 12 m \\ Volume of conical pit = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times 12 \\ = 11 \times 3.5 = 38.5 m^{3} \\ = 38.5 kilolitres [∵ 1 m^{3} = 1 kilolitre] \\ Thus, the capacity of conical pit is 38.5 kilolitres . \end{array}$

Q.54 The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find:

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone.

Ans.

$\begin{matrix} Diameter of a right circular cone = 28 c m \\ Radius of a right circular cone = 14 c m \\ (i) Volume of a right circular cone = 9856 c m^{3} \\ \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times h = 9856 c m^{3} \\ h = \frac{9856 \times 3}{22 \times 2 \times 14} \\ = 48 c m \\ (i i) Slant height of the cone (l) = \sqrt{r^{2} + h^{2}} \\ = \sqrt{14^{2} + 48^{2}} \\ = \sqrt{2500} \\ = 50 c m \\ (i i i) Curved surface area of cone = π r l \\ = \frac{22}{7} \times 14 \times 50 \\ = 2200 {cm}^{2} \end{matrix}$

Thus, the curved surface area of cone is 2200 cm².

Q.55 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained..

Ans.

$\begin{array}{l} In ΔABC, let BC = 5 cm, AB = 12 cm and AC = 13 c m \\ When ΔABC is revolved around AB, a cone is obtained, \\ whose radius is 5 cm, height is 12 cm and slant height is 13 cm. \\ So, Volume of cone = \frac{1}{3} {πr}^{2} h \\ = \frac{1}{3} π {(5)}^{2} \times 12 \\ = 100 π {cm}^{2} \\ Thus, the volume of cone is 100 π {cm}^{2} . \end{array}$

Q.56 If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Ans.

$\begin{array}{l} When Δ ABC is revolved around BC, then radius of cone is \\ 12 cm, height is 5 cm and slant height is 13 cm . \\ ∴ Volume of cone = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} π {(12)}^{2} \times 5 \\ = \frac{1}{3} π \times 12 \times 12 \times 5 \\ = 240 π c m^{3} \\ Ratio of volumes of the two solids obtained on revolving triangle \\ = \frac{100 π}{240 π} \\ = 5 : 12 \end{array}$

Q.57 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Ans.

$\begin{array}{l} Diameter of conical heap = 10.5 m \\ Radius of conical heap = \frac{10.5}{2} m \\ Height of conical heap = 3 m \\ Volume of conical heap = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times \frac{10.5}{2} \times \frac{10.5}{2} \times 3 \\ = \frac{22}{7} \times \frac{10515}{20} \times \frac{105}{20} \\ = \frac{11}{10} \times 15 \times \frac{21}{4} \\ = 86.625 c m^{3} \\ Slant height of conical heap \\ = \sqrt{3^{2} + {(\frac{10.5}{2})}^{2}} \\ = \sqrt{9 + \frac{110.25}{4}} \\ = \sqrt{\frac{36 + 110.25}{4}} \\ = \sqrt{\frac{146.25}{4}} \\ = \frac{12.1}{2} \\ = 6.05 m \\ Curved surface area of conical heap \\ = π r l \\ = \frac{22}{7} \times \frac{10.5}{2} \times 6.05 \\ = 99.825 m^{2} \\ Thus, the area of canvas required is 99 {.825 m}^{2} . \end{array}$

Q.58 Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m.

Ans.

$\begin{array}{l} Volume of sphere = \frac{4}{3} π r^{3} \\ (i) W h e n radius = 7 cm \\ Volume of sphere = \frac{4}{3} π {(7)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \\ = 1437.33 c m^{3} \\ (i i) When radius = 0 .63 cm \\ Volume of sphere = \frac{4}{3} π {(0 .63)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 0 .63 \times 0 .63 \times 0 .63 \\ = 4 \times 22 \times 0 .03 \times 0 .63 \times 0 .63 \\ = 1.047816 c m^{3} \\ = 1.05 c m^{3} (a p p r o x .) \end{array}$

Q.59 Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m, when the solid spherical ball is placed in a large tank containing water.

Ans.

$\begin{array}{l} Volume of sphere = \frac{4}{3} π r^{3} \\ Water displaced by a spherical ball is equal to the volume of the sphere. \\ (i) When diameter of sphere = 28 cm \\ Radius of sphere = 14 c m \\ Volume of sphere = \frac{4}{3} π {(14)}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 \\ = 11498.64 c m^{3} \\ (ii) When diameter of sphere = 0. 21 m \\ R a d i u s of sphere = \frac{0.21}{2} m \\ Volume of sphere = \frac{4}{3} π {(\frac{0.21}{2})}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{0.21}{2} \times \frac{0.21}{2} \times \frac{0.21}{2} \\ = 11 \times 0 .01 \times 0 .21 \times 0 .21 \\ = 0.004851 m^{3} \end{array}$

Q.60 The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Ans.

$\begin{array}{l} D i a m e t e r of a metallic ball = 4.2 c m \\ Radius of a metallic ball = 2.1 c m \\ Volume of metallic ball = \frac{4}{3} π r^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \\ = 4 \times 22 \times 0.1 \times 2.1 \times 2.1 \\ = 38.808 c m^{3} \\ Density of metal = 8.9 g p e r c m^{3} \\ Mass of metallic ball = V o l u m e \times D e n s i t y \\ = 38.808 \times 8.9 \\ = 345.39 grams . \end{array}$

Q.61 The diameter of the moon is approximately one-fourth of the diameter of the earth..What fraction of the volume of the earth is the volume of the moon?

Ans.

$\begin{matrix} Radius of earth (R_{e}) = \frac{Diameter of earth}{2} \\ = \frac{D_{e}}{2} \\ Radius of moon (R_{m}) = \frac{\frac{1}{4} \times D_{e}}{2} \\ = \frac{D_{e}}{8} \\ \frac{Volume of moon}{Volume of earth} = \frac{\frac{4}{3} π {(R_{m})}^{3}}{\frac{4}{3} π {(R_{e})}^{3}} \\ = \frac{\frac{4}{3} π {(\frac{D_{e}}{8})}^{3}}{\frac{4}{3} π {(\frac{D_{e}}{2})}^{3}} \\ = \frac{\frac{1}{8} \times \frac{1}{8} \times \frac{1}{8}}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} \\ = \frac{1}{64} \\ Volume of moon = \frac{1}{64} \times Volume of earth \end{matrix}$

Q.62 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Ans.

$\begin{array}{l} Diameter of hemispherical bowl = 10.5 c m \\ Radius of hemispherical bowl = \frac{10.5}{2} c m \\ Volume of hemispherical bowl = \frac{2}{3} π r^{3} \\ = \frac{2}{3} \times \frac{22}{7} \times \frac{10.5}{2} \times \frac{10.5}{2} \times \frac{10.5}{2} \\ = 11 \times \frac{5}{20} \times \frac{105}{10} \times \frac{105}{10} \\ = 11 \times \frac{1}{4} \times \frac{21}{2} \times \frac{21}{2} \\ = 303.1875 c m^{3} \\ = \frac{303.1875}{1000} l i t r e s (a p p r o x .) \\ [∵ 1 c m^{3} = \frac{1}{1000} l i t r e s] \\ = 0.3031875 l i t r e s \\ Capacity of hemispherical bowl = 0.303 l i t r e s . \end{array}$

Q.63 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Ans.

$\begin{array}{l} The inner radius of hemispherical tank (r_{2}) = 1 m \\ Thickness of hemispherical tank = 1 c m \\ = 0.01 m \\ Oute r radius of spherical tank (r_{1}) = 1 + 0.01 \\ = 1.01 m \\ Volume of iron used to make the tank = \frac{2}{3} π (r_{1}^{3} - r_{2}^{3}) \\ = \frac{2}{3} \times \frac{22}{7} {{(1.01)}^{3} - {(1)}^{3}} \\ = \frac{44}{21} (1.030301 - 1) \\ = \frac{44}{21} (0.030301) \\ = 0.06348 m^{3} \\ Thus, the volume of iron used to make the tank is 0.06348 m^{3} . \end{array}$

Q.64 Find the volume of a sphere whose surface area is 154 cm².

Ans.

$\begin{array}{l} Surface area of a sphere = 154 c m^{2} \\ 4 \times \frac{22}{7} \times r^{2} = 154 c m^{2} \\ r^{2} = \frac{154 \times 7}{4 \times 22} \\ = \frac{7 \times 7}{4} \\ r = \sqrt{\frac{7 \times 7}{4}} \\ = \frac{7}{2} c m \\ Volume of a sphere = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \\ = \frac{11 \times 49}{3} = \frac{539}{3} \\ = 179 \frac{2}{3} c m^{3} \\ Thus, volume of sphere is 179 \frac{2}{3} c m^{3} . \end{array}$

Q.65 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 498.96. If the cost of white-washing is ₹ 2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Ans.

$\begin{array}{l} (i) \\ Cost of white-washing the hemisphere = ₹ 498 .96 \\ Rate of white-washing the hemisphere = ₹ 2 .00 / m^{2} \\ Therefore, Surface area of the hemisphere = \frac{498 .96}{2} \\ = 249 .48 m^{2} \\ Thus, inside surface area of the dome is 249 .48 m^{2} . \\ (ii) ∵ Surface area of the hemisphere = 249 .48 m^{2} \\ 2 {πr}^{2} = 249 .48 m^{2} \\ 2 \times \frac{22}{7} \times r^{2} = 249 .48 m^{2} \\ r = \sqrt{\frac{249 .48 \times 7}{2 \times 22}} \\ r = 6 .3 m \\ Volume of air inside the tomb = \frac{4}{3} {πr}^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times 6 .3 \times 6 .3 \times 6 .3 \\ = 88 \times 0 .3 \times 6 .3 \times 6 .3 \\ = 1047 .816 m^{3} \\ Thus, the volume of air inside the tomb is 1047 .816 m^{3} . \end{array}$

Q.66 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r ′ of the new sphere,
(ii) ratio of S and S′.

Ans.

$\begin{array}{l} (i) Volume of new sphere = 27 \times volume of solid iron sphere \\ \frac{4}{3} π {(r ‘)}^{3} = 27 \times \frac{4}{3} {πr}^{3} \\ \Rightarrow r ‘ = \sqrt{27 r^{3}} \\ \Rightarrow r ‘ = 3 r \\ Thus, the radius (r ‘) of new sphere is 3r. \\ (ii) Ratio of S and S’ = \frac{4 {πr}^{2}}{4 πr ‘^{2}} \\ = \frac{r^{2}}{{(3 r)}^{2}} [∵ r ‘ = 3 r] \\ = \frac{1}{9} \\ Thus, ration of S and S’ is 1:9. \end{array}$

Q.67 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine material (in mm³) is needed to fill this capsule?

Ans.

$\begin{array}{l} Diameter of spherical capsule = 3 . 5 mm \\ Radius of spherical capsule = \frac{3.5}{2} mm \\ Volume of spherical capsule = \frac{4}{3} π r^{3} \\ = \frac{4}{3} \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \times \frac{3.5}{2} \\ = \frac{11}{3} \times 0.5 \times 3.5 \times 3.5 \\ = 22.458 m m^{3} \\ = 22.46 m m^{3} (A p p r o x .) \\ Thus, 22.46 m m^{3} medicine materials are required to fill the capsule . \end{array}$

Q.68 A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see Figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf.

Ans.

$\begin{array}{l} Outer surface area = 2 (110 + 85) \times 25 + 85 \times 110 \\ = 2 \times 195 \times 25 + 9350 \\ = 19100 {cm}^{2} \\ Area vertical edges = 110 \times 5 + 110 \times 5 \\ = 1100 {cm}^{2} \\ Area horizontal edges \\ = 75 \times 5 + 75 \times 5 + 75 \times 5 + 75 \times 5 \\ = 75 (5 + 5 + 5 + 5) \\ = 75 \times 20 \\ = 1500 {cm}^{2} \\ Total external area = 19100 {cm}^{2} + 1100 {cm}^{2} + 1500 {cm}^{2} \\ = 21700 {cm}^{2} \\ Internal length of each rack of bookshelf \\ = 85 - 5 - 5 \\ = 75 cm \\ Internal breadth of each rack of bookshelf \\ = \frac{110 - 20}{3} \\ = 30 cm \\ Internal depth of each rack of bookshelf \\ = 25 - 5 \\ = 20 cm \\ Surface area of a bookshelf \\ = 2 (75 + 30) \times 20 + 75 \times 30 \\ = 2 \times 105 \times 20 + 2250 \end{array}$ $\begin{array}{l} = 4200 + 2250 \\ = 6450 {cm}^{2} \\ Surface area of three bookshelves \\ = 3 \times 6450 {cm}^{2} \\ = 19350 {cm}^{2} \\ Rate of polishing external surface of bookshelf \\ = ₹ \frac{20}{100} \\ = ₹ 0 .20 {per m}^{2} \\ Rate of painting internal surface of bookshelf \\ = ₹ \frac{10}{100} \\ = ₹ 0 .10 {per m}^{2} \\ Total cost of polishing and painting the bookshelf \\ = 0 .20 \times 21700 + 0 .10 \times 19350 \\ = 4340 + 1935 \\ = ₹ 6275 \\ Thus, total expenses required for polishing and painting the \\ surface of the bookshelf is ₹ 6275. \end{array}$

Q.69 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Ans.

$\begin{array}{l} Diameter of wooden sphere = 21 c m \\ Radius of wooden sphere = \frac{21}{2} c m \\ Radius of each cylinder = 1.5 c m \\ Height of each cylinder = 7 c m \\ Surface area of sphere = 4 π r^{2} \\ = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \\ = 22 \times 3 \times 21 \\ = 1386 c m^{2} \\ Area of top of cylinder = π r^{2} \\ = \frac{22}{7} \times 1.5 \times 1.5 \\ = 7.07 c m^{2} \\ Surface area of sphere to be painted \\ = 1386 - 7.07 \\ = 1378.93 c m^{2} \\ Curved surface area of cylinder \\ = 2 \times \frac{22}{7} \times 1.5 \times 7 \end{array}$ $\begin{array}{l} = 66 c m^{2} \\ Cost of silver painting on sphere \\ = \frac{25}{100} \\ = R s . 0.25 \\ Cost of black painting on cylinder \\ = R s . \frac{5}{100} \\ = R s . 0.05 \\ C o s t of painting of one sphere and cylinder \\ = R s . 0.25 \times 1378.93 + R s . 0.05 \times 66 \\ = R s . (344.7325 + 3.3) \\ = R s .348.0325 \\ Cost of painting of eight combinatin of sphere and cylinder \\ = 8 \times R s .348.0325 \\ = R s .2784.26 \end{array}$

Q.70 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Ans.

$\begin{array}{l} L e t diameter of sphere be 2r . \\ Curved surface area of sphere = 4 π r^{2} \\ Decreased radius of sphere = r - 25 % of r \\ = r - \frac{25}{100} \times r \end{array}$ $\begin{array}{l} = r - \frac{r}{4} \\ = \frac{3 r}{4} \\ New curved surface area of sphere \\ = 4 π {(\frac{3 r}{4})}^{2} \\ = \frac{9}{4} π r^{2} \\ Percentage of decreased curved surface area \\ = \frac{4 π r^{2} - \frac{9}{4} π r^{2}}{4 π r^{2}} \times 100 % \\ = \frac{\frac{7}{4} π r^{2}}{4 π r^{2}} \times 100 % \\ = \frac{7}{16} \times 100 % \\ = 43.75 % \\ Thus, the 43 .75% curved surface area of sphere is decreased . \end{array}$

Q.71 Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find

(i) radius of the base
(ii) total surface area of the cone.

Ans.

(i) Curved surface area of cone = 308 cm²
πrl = 308 cm²
(22/7) x r x 14 = 308 cm²
r = (308/14) x (7/ 22)
= 7 cm
Thus, the radius of the base of cone is 7 cm.

(ii) Total surface area of the cone = πr(l + r)
= (22/7) x 7(14 + 7)
= 22 x 21

= 462 cm²
Thus, total surface area of the cone is 462 cm².

Q.72 What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14).

Ans.

$\begin{array}{l} Height of conical tent = 8 m \\ Base radius of conical tent = 6 m \\ Slant height of conical tent = \sqrt{8^{2} + 6^{2}} \\ = \sqrt{100} \\ = 10 m \\ surface area of conical tent = π r l \\ = 3.14 \times 6 \times 10 \\ = 188.4 m^{2} \\ Width of tarpaulin = 3 m \\ Required length of tarpaulin = \frac{188.4}{3} \\ = 62.8 m \\ Wastage in cutting of tarpaulin \\ = 20 cm \\ = \frac{20}{100} m \\ = 0.2 m \\ Required tarpaulin for tent = 62.8 + 0.2 \\ = 63 m \\ Thus, the length of the required tarpaulin will be 63 m . \end{array}$

Q.73 If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find

(i) radius of its base

(ii) its volume.

(Use π = 3.14)

Ans.

$\begin{array}{l} (i) The height of cylinder = 5 cm \\ The lateral surface area of cylinder \\ = 94.2 c m^{2} \\ 2 π r h = 94.2 {cm}^{2} \\ \Rightarrow 2 \times 3.14 \times r \times 5 = 94.2 {cm}^{2} \\ \Rightarrow r = \frac{94.2}{2 \times 3.14 \times 5} \\ = 3 cm \\ Thus, the radius of cylinder is 3 cm . \\ (i i) Volume of cylinder = π r^{2} h \\ = 3.14 \times 3 \times 3 \times 5 \\ = 141.3 c m^{3} \\ Thus, the volume of cylinder is 141 {.2 cm}^{3} . \end{array}$

Q.74 The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Ans.

$\begin{array}{l} Height of the cone = 15 c m \\ Volume of the cone = 1570 c m^{3} \\ \frac{1}{3} π r^{2} h = 1570 c m^{3} \\ \Rightarrow \frac{1}{3} \times 3.14 \times r^{2} \times 15 = 1570 c m^{3} \\ \Rightarrow r^{2} = \frac{1570}{3.14 \times 5} \\ \Rightarrow r = 10 \\ Thus, the radius of the base is 10 cm . \end{array}$

Q.75 If the volume of a right circular cone of height 9 cm is 48 π cm³, find the diameter of its base.

Ans.

$\begin{array}{l} Heigh t of the cone = 9 c m \\ Volume of the cone = 48 π c m^{3} \\ \frac{1}{3} π r^{2} h = 48 π c m^{3} \\ \Rightarrow \frac{1}{3} \times r^{2} \times 9 = 48 c m^{3} \\ \Rightarrow r^{2} = \frac{48}{3} \\ \Rightarrow = 16 \\ \Rightarrow r = 4 \\ diameter = 2 \times 4 \\ = 8 c m \\ Thus, the diameter of the base is 8 cm . \end{array}$

Please register to view this section

FAQs (Frequently Asked Questions)

NCERT Class 9 Mathematics Chapter 13 has detailed solutions to each exercise and every question of the chapter.. As a result, the formulas included in the solutions can be helpful for students. Students are advised to read and revise these formulas again and again in order to be thorough with them and to score well in this chapter.

You can refer to NCERT study material and NCERT Solutions for Class 9 Mathematics Chapter 13 on the Extramarks’ website, a trusted source by students and teachers.

To get the NCERT Solutions for Class 9 Mathematics Chapter 13, you need to follow the following steps:

Search for the Extramarks’ website on your web browser
Click on the list of contents tab in the upper left corner of the website
Click on the study material option from the list of contents
Select CBSE and then select Class IX
Select Mathematics
Sign up if you don’t have an existing account, or log in if you already have an account
Now select NCERT Solutions for Class 9 Mathematics Chapter 13.

In this way, you can get all the NCERT Class 9 Mathematics-related study material from the Extramarks’ website.

NCERT Solutions for Class 9 Maths Chapter 13- Surface areas and volumes

Key Topics Covered In NCERT Solutions for Class 9 Maths Chapter 13

Introduction

Surface area of cuboid and cube

NCERT Solutions for Class 9 Maths Chapter 13: Exercise & Solutions

NCERT Exemplar for Class 9 Maths

Key Features of NCERT Solutions for Class 9 Maths Chapter 13

Please register to view this section

FAQs (Frequently Asked Questions)

Company

Terms & Conditions

Resources

Teachers

NCERT Solutions for Class 9 Maths Chapter 13- Surface areas and volumes

Key Topics Covered In NCERT Solutions for Class 9 Maths Chapter 13

Introduction

Surface area of cuboid and cube

NCERT Solutions for Class 9 Maths Chapter 13: Exercise & Solutions

NCERT Exemplar for Class 9 Maths

Key Features of NCERT Solutions for Class 9 Maths Chapter 13

Please register to view this section

FAQs (Frequently Asked Questions)

1. How can I learn all the formulas included in NCERT Class 9 Mathematics Chapter 13 thoroughly?

2. How can I get NCERT Solutions for Class 9 Mathematics Chapter 13?

Company

Terms & Conditions

Resources

Teachers